RBSE Solutions Class 12 Physics Chapter 14 Atomic Physics

RBSE Class 12 Physics Chapter 14 Multiple Choice Type Questions

 

Question 1. The ground state energy of hydrogen atom is -13.6 eV. What will be its energy for n = 5 level?
(a) -0.54 eV
(b) -0.85 eV
(c) -5.4 eV
(d) -2.72 eV
Answer: (a) -0.54 eV
In simple words: The energy of an electron in a hydrogen atom changes depending on its orbit (n level). For the fifth orbit, the energy becomes -0.54 eV, which is less negative than the ground state, showing it's less tightly bound.

🎯 Exam Tip: Remember the formula for energy levels in a hydrogen atom: \( E_n = \frac{-13.6 \text{ eV}}{n^2} \). Just plug in the value of 'n' to find the energy for any given level.

 

Question 2. The energy of hydrogen atom in nth orbit is \( E_n = -\frac{13.6}{n^2} \), eV. Calculate the energy necessary to send electron from first orbit to second orbit:
(a) 10.2 eV
(b) 12.1 eV
(c) 13.0 eV
(d) 3.4 eV
Answer: (a) 10.2 eV
The energy of hydrogen atom in nth orbit is given by \( E_n = -\frac{13.6}{n^2} \text{ eV} \)
For the first orbit, \( n = 1 \):
\( E_1 = -\frac{13.6}{(1)^2} = -13.6 \text{ eV} \)
For the second orbit, \( n = 2 \):
\( E_2 = -\frac{13.6}{(2)^2} = -\frac{13.6}{4} = -3.4 \text{ eV} \)
The energy necessary to send an electron from the first orbit to the second orbit is the difference in energy between these two levels.
\( \Delta E = E_2 - E_1 \)
\( \Delta E = -3.4 - (-13.6) \)
\( \Delta E = -3.4 + 13.6 \)
\( \Delta E = 10.2 \text{ eV} \)
In simple words: To move an electron from its first energy level to the second, you need to provide 10.2 eV of energy. This is because the electron needs to absorb energy to jump to a higher, less negative energy state.

🎯 Exam Tip: Always pay attention to the signs when calculating energy differences; a negative sign indicates binding energy, and the electron absorbs energy to move to a higher state.

 

Question 3. In hydrogen atom if electron transit from 3rd orbit to 2nd orbit, then wavelength of the emitted radiation will be:
(c)
Answer: (c) \( \lambda = \frac{36}{5R} \)
The wavelength of emitted radiation during a transition in a hydrogen atom is given by the Rydberg formula:
\( \frac{1}{\lambda} = R \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right] \)
Here, the electron transitions from the 3rd orbit (\( n_2 = 3 \)) to the 2nd orbit (\( n_1 = 2 \)).
So, \( n_1 = 2 \), \( n_2 = 3 \).
Now, put these values into the formula:
\( \frac{1}{\lambda} = R \left[\frac{1}{2^2} - \frac{1}{3^2}\right] \)

\( \frac{1}{\lambda} = R \left[\frac{1}{4} - \frac{1}{9}\right] \)

\( \frac{1}{\lambda} = R \left[\frac{9 - 4}{36}\right] \)

\( \frac{1}{\lambda} = R \left[\frac{5}{36}\right] \)

\( \frac{1}{\lambda} = \frac{5R}{36} \)
Now, to find \( \lambda \), simply invert the equation:
\( \lambda = \frac{36}{5R} \)
In simple words: When an electron moves from a higher energy level (like the 3rd orbit) to a lower one (like the 2nd orbit), it releases energy as light. This light has a specific wavelength, which can be calculated using a special formula related to the Rydberg constant (R) and the orbit numbers.

🎯 Exam Tip: Always remember that \( n_1 \) refers to the lower energy level (final orbit) and \( n_2 \) refers to the higher energy level (initial orbit) in the Rydberg formula for emission.

 

Question 4. In which part of the electromagnetic spectrum the Lyman series of hydrogen is found?
(a) ultraviolet
(b) infrared
(c) visible
(d) X-ray region
Answer: (a) ultraviolet
In simple words: The Lyman series is a set of spectral lines emitted by hydrogen atoms when electrons jump down to the first energy level. These particular light waves have very high energy and fall into the ultraviolet part of the electromagnetic spectrum, which is invisible to the human eye.

🎯 Exam Tip: Remember that transitions to \( n=1 \) (Lyman series) are in the ultraviolet region, \( n=2 \) (Balmer series) include visible light, and transitions to higher principal quantum numbers are in the infrared region.

 

Question 5. A hydrogen atom is excited to energy level n = 4 Calculate the emitted spectrum lines:
(a) 2
(b) 3
(c) 4
(d) 6
Answer: (d) 6
When an electron in a hydrogen atom is excited to an energy level \( n \), the total number of possible spectral lines (emitted when it de-excites to lower levels) can be calculated using the formula:
\( N_E = \frac{n(n-1)}{2} \)
Given that the atom is excited to energy level \( n = 4 \).
Substitute \( n = 4 \) into the formula:
\( N_E = \frac{4(4-1)}{2} \)
\( N_E = \frac{4(3)}{2} \)
\( N_E = \frac{12}{2} \)
\( N_E = 6 \)
So, 6 different spectrum lines can be emitted when the electron de-excites from \( n=4 \).
In simple words: If a hydrogen atom's electron gets enough energy to reach the 4th energy level, it can then fall back down in several different steps, creating different light colors. The total number of these possible light emissions from the 4th level is 6.

🎯 Exam Tip: The formula \( \frac{n(n-1)}{2} \) calculates the maximum number of distinct spectral lines when an electron transitions from the \( n^{th} \) excited state down to any lower state, including the ground state.

 

Question 7. In excited hydrogen atom according to Bohr's theory if angular momentum is \( \left(\frac{2 h}{2 \pi}\right) \), then its energy will be:
(a) -13.6 eV
(b) -13.4 eV
(c) -3.4 eV
(d) -12.8 eV
Answer: (c) -3.4 eV
According to Bohr's second postulate, the angular momentum (L) of an electron in an orbit is quantized:
\( L = \frac{nh}{2\pi} \)
Given that the angular momentum \( L = \frac{2h}{2\pi} \).
By comparing this with the formula, we can see that \( n = 2 \).
Now, we need to find the energy of the electron for \( n=2 \) in a hydrogen atom. The energy formula for a hydrogen atom is:
\( E_n = -\frac{13.6}{n^2} \text{ eV} \)
Substitute \( n = 2 \) into the energy formula:
\( E_2 = -\frac{13.6}{(2)^2} \text{ eV} \)
\( E_2 = -\frac{13.6}{4} \text{ eV} \)
\( E_2 = -3.4 \text{ eV} \)
In simple words: Bohr's theory states that an electron's angular momentum can only take certain values, which tells us its orbit number (n). If the angular momentum is given as \( \frac{2h}{2\pi} \), it means the electron is in the second orbit (\( n=2 \)). For the second orbit of a hydrogen atom, the electron's energy is -3.4 eV.

🎯 Exam Tip: Remember to first determine the principal quantum number 'n' from the given angular momentum before calculating the energy. Always double-check the 'n' value and its square in the energy formula.

 

Question 9. In the following constants which is equal to ground states for all hydrogen similar ions?
(a) orbital speed of electron
(b) radius of orbit
(c) angular momentum of electron
(d) energy of the atom
Answer: (c) angular momentum of electron
In simple words: For all hydrogen-like ions (atoms with only one electron, like He+, Li++), the angular momentum of the electron in its lowest energy state (ground state) is always the same. This is a fundamental principle in Bohr's model, where angular momentum is quantized.

🎯 Exam Tip: Bohr's quantization condition for angular momentum \( L = \frac{nh}{2\pi} \) holds for all single-electron systems. For the ground state (\( n=1 \)), L is always \( \frac{h}{2\pi} \), regardless of the atomic number (Z).

 

Question 10. The ground state energy of an hydrogen similar ion is -54.4 eV. It may be:
(a) He+
(b) Li ++
(c) Deuterium+
(d) Be+++
Answer: (a) He+
The energy of a hydrogen-like ion in the nth orbit is given by the formula:
\( E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \)
For the ground state, \( n=1 \).
So, \( E_1 = -\frac{13.6 Z^2}{(1)^2} = -13.6 Z^2 \text{ eV} \)
Given that the ground state energy is \( -54.4 \text{ eV} \).
So, \( -54.4 = -13.6 Z^2 \)
Divide both sides by -13.6:
\( Z^2 = \frac{-54.4}{-13.6} \)
\( Z^2 = 4 \)
Now, take the square root to find Z:
\( Z = \sqrt{4} \)
\( Z = 2 \)
An ion with atomic number \( Z = 2 \) is Helium (He). Since it is a hydrogen-like ion (meaning it has only one electron), it must be \( \text{He}^+ \).
In simple words: By using the energy formula for hydrogen-like atoms and the given ground state energy, we can calculate the atomic number (Z). A Z value of 2 means the atom is Helium, and since it's a hydrogen-like ion, it has lost one electron, making it \( \text{He}^+ \).

🎯 Exam Tip: Remember that "hydrogen-like ion" means a species with only one electron, regardless of its atomic number (Z). Use the \( Z^2 \) factor in the energy formula carefully.

 

Question 11. On increase in the value of principal quantum number in hydrogen atom, the potential energy of the atom:
(a) decreases
(b) increases
(c) remains same
(d) potential energy increases and decreases alternatively
Answer: (a) decreases
In simple words: As the principal quantum number (n) increases, the electron moves to orbits farther from the nucleus. This makes the potential energy of the atom decrease (become less negative, or effectively increase in value), as the electron is less tightly bound to the nucleus.

🎯 Exam Tip: For bound systems, potential energy is negative. Moving to higher orbits means the electron is less bound, so its potential energy becomes less negative (increases towards zero), which is a decrease in magnitude.

 

Question 12. Hydrogen atom can transit from n = 4 to n = 1 state. Then what will be projection momentum of hydrogen atom (in eV/c)?
(a) 13.60
(b) 12.75
(c) 0.85
(d) 22.1
Answer: (b) 12.75
When a hydrogen atom transitions from \( n=4 \) to \( n=1 \), the energy released is:
\( \Delta E = E_4 - E_1 \)
Using the energy formula \( E_n = -\frac{13.6}{n^2} \text{ eV} \):
\( E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \text{ eV} \)
\( E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} \)
\( \Delta E = -0.85 - (-13.6) = -0.85 + 13.6 = 12.75 \text{ eV} \)
The energy released is 12.75 eV. This energy is carried by the emitted photon. The momentum of a photon is given by \( p = \frac{E}{c} \).
So, the projection momentum of the hydrogen atom after emitting this photon will be equal to the momentum of the photon.
\( p = \frac{12.75 \text{ eV}}{c} \)
The value is 12.75 in units of eV/c.
In simple words: When an electron in a hydrogen atom drops from a higher energy level (n=4) to a lower one (n=1), it releases energy. This energy, 12.75 eV, is carried away by a photon. The atom recoils with a momentum equal to the momentum of this emitted photon, which is 12.75 eV/c.

🎯 Exam Tip: The projection momentum of the atom after photon emission is numerically equal to the photon's momentum, \( \frac{\Delta E}{c} \). Remember to calculate \( \Delta E \) accurately first.

 

Question 13. In the nth class of hydrogen atom (angular momentum L), the magnetic moment due to orbital motion of electron is:
(a) \( \frac{neL}{2m} \)
(b) \( -\frac{eL}{2m} \)
(c) \( -\frac{eL}{2mn} \)
(d) \( -\frac{eLm}{n} \)
Answer: (b) \( -\frac{eL}{2m} \)
The magnetic dipole moment \( \mu \) due to the orbital motion of an electron is given by:
\( \mu = I A \)
where I is the current and A is the area of the orbit.
The current \( I = \frac{e}{T} \), where T is the time period of revolution.
The area \( A = \pi r^2 \).
Also, \( T = \frac{2\pi r}{v} \), where v is the orbital speed.
So, \( I = \frac{e}{\left(\frac{2\pi r}{v}\right)} = \frac{ev}{2\pi r} \)
Then, \( \mu = \left(\frac{ev}{2\pi r}\right) (\pi r^2) \)
\( \mu = \frac{evr}{2} \)
Multiply and divide by mass \( m \):
\( \mu = \frac{emvr}{2m} \)
We know that angular momentum \( L = mvr \).
So, \( \mu = \frac{eL}{2m} \)
The negative sign indicates that the magnetic moment is opposite to the direction of angular momentum due to the negative charge of the electron.
Thus, \( \mu = -\frac{eL}{2m} \)
In simple words: An electron moving in an orbit creates a tiny current, which then produces a magnetic field. This magnetic effect is called magnetic moment. The formula for this magnetic moment is directly linked to the electron's charge (e), its angular momentum (L), and its mass (m), with a negative sign showing the directions are opposite.

🎯 Exam Tip: Remember that the orbital magnetic moment is proportional to the angular momentum. The negative sign is crucial as it indicates the opposite direction due to the electron's negative charge.

 

Question 14. If an electron transitions from ground state to first excited state, then the increase in its angular momentum is:
(b) \( 1.05 \times 10^{-34} \) J-s
Answer: (b) \( 1.05 \times 10^{-34} \) J-s
According to Bohr's theory, angular momentum is quantized:
\( L = \frac{nh}{2\pi} \)
The ground state corresponds to \( n_1 = 1 \).
The first excited state corresponds to \( n_2 = 2 \).
The initial angular momentum in the ground state is:
\( L_1 = \frac{1 \cdot h}{2\pi} = \frac{h}{2\pi} \)
The final angular momentum in the first excited state is:
\( L_2 = \frac{2 \cdot h}{2\pi} = \frac{h}{\pi} \)
The increase in angular momentum is \( \Delta L = L_2 - L_1 \).
\( \Delta L = \frac{h}{\pi} - \frac{h}{2\pi} \)
\( \Delta L = \frac{2h - h}{2\pi} = \frac{h}{2\pi} \)
Now, substitute the value of Planck's constant \( h = 6.63 \times 10^{-34} \text{ J-s} \) and \( \pi = 3.14 \).
\( \Delta L = \frac{6.63 \times 10^{-34}}{2 \times 3.14} \)
\( \Delta L = \frac{6.63 \times 10^{-34}}{6.28} \)
\( \Delta L \approx 1.055 \times 10^{-34} \text{ J-s} \)
Rounding to two decimal places, this is \( 1.05 \times 10^{-34} \text{ J-s} \).
In simple words: When an electron jumps from its lowest energy state (ground state, n=1) to the next higher one (first excited state, n=2), its angular momentum increases. This increase is a specific value determined by Planck's constant and pi, and it is \( 1.05 \times 10^{-34} \) J-s.

🎯 Exam Tip: Always remember that the ground state is \( n=1 \), the first excited state is \( n=2 \), the second excited state is \( n=3 \), and so on. Use the quantized angular momentum formula to find the change.

RBSE Class 12 Physics Chapter 14 Very Short Answer Type Questions

 

Question 1. Most of the positive charge of the atom is centralised in a small centre. By which experiment it was proved?
Answer: Rutherford a-ray scattering experiment.
In simple words: The experiment where Rutherford shot tiny alpha particles at a thin gold foil proved that almost all the positive charge and mass of an atom are packed into a very tiny central part called the nucleus.

🎯 Exam Tip: When asked about the discovery of the atomic nucleus, always mention Rutherford's alpha-ray scattering experiment as the key evidence.

 

Question 2. Write two shortcomings of Rutherford model.
Answer: The two main shortcomings of Rutherford's atomic model were:
1. It failed to explain the stability of an atom, as orbiting electrons should continuously radiate energy and eventually spiral into the nucleus.
2. It could not explain the discrete line spectrum of atoms; instead, it predicted a continuous spectrum.
In simple words: Rutherford's model had two problems: it couldn't explain why atoms are stable and don't collapse, and it couldn't explain why atoms emit specific colors of light instead of all colors.

🎯 Exam Tip: The two biggest criticisms of Rutherford's model are its inability to account for atomic stability and the observation of discrete atomic spectra.

 

Question 3. If angular momentum of the electron in hydrogen atom is \( \frac{h}{\pi} \), then in which class it will be n situated?
Answer: Based on Bohr's quantization condition, \( L = n \frac{h}{2\pi} \).
Given the angular momentum \( L = \frac{h}{\pi} \).
Comparing the two, we have \( n \frac{h}{2\pi} = \frac{h}{\pi} \).
We can cancel \( \frac{h}{\pi} \) from both sides, which gives \( \frac{n}{2} = 1 \).
Therefore, \( n = 2 \). The electron will be situated in the second orbit (n=2). This principle helps explain the specific energy levels of electrons.
In simple words: If an electron's angular momentum is \( \frac{h}{\pi} \), it means the electron is in the second energy level or orbit (n=2) of the hydrogen atom.

🎯 Exam Tip: To find the principal quantum number 'n' from angular momentum, always compare the given angular momentum value with Bohr's quantization formula \( L = \frac{nh}{2\pi} \).

 

Question 4. Lyman series of hydrogen lies in which part of electromagnetic spectrum?
Answer: In ultraviolet region.
In simple words: The Lyman series is a set of specific light emissions from hydrogen. These emissions fall into the ultraviolet part of the light spectrum.

🎯 Exam Tip: Remember that the Lyman series involves electron transitions to the ground state (\( n=1 \)) and is characterized by high energy, hence its placement in the ultraviolet spectrum.

 

Question 5. In first Bohr orbit of hydrogen similar atom the energy of electron is -27.2 eV. What will be its energy in third Bohr orbit?
Answer: The energy of an electron in a hydrogen-like atom in the \( n^{th} \) orbit is given by:
\( E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \)
Given that for the first Bohr orbit (\( n=1 \)), the energy \( E_1 = -27.2 \text{ eV} \).
So, \( -27.2 = -\frac{13.6 Z^2}{(1)^2} \)
\( -27.2 = -13.6 Z^2 \)
\( Z^2 = \frac{-27.2}{-13.6} = 2 \)
Now we need to find the energy in the third Bohr orbit (\( n=3 \)) for this atom. We will use the calculated \( Z^2 = 2 \).
\( E_3 = -\frac{13.6 Z^2}{(3)^2} \)
\( E_3 = -\frac{13.6 \times 2}{9} \)
\( E_3 = -\frac{27.2}{9} \)
\( E_3 \approx -3.02 \text{ eV} \)
In simple words: We first used the given energy for the first orbit to find a special value (Z-squared) for this specific atom, which turned out to be 2. Then, using this value, we calculated the energy for the third orbit, which is about -3.02 eV.

🎯 Exam Tip: When dealing with "hydrogen-like atoms," always use the given energy for a known orbit (usually the ground state) to first determine \( Z^2 \) for that particular atom before calculating energy for other orbits.

 

Question 6. What will be the relation of radius of different classes of hydrogen atom?
Answer: The radius of the \( n^{th} \) Bohr orbit for a hydrogen atom is given by:
\( r_n = \frac{\epsilon_0 h^2}{\pi m e^2} n^2 \)
This can be written as \( r_n = r_1 n^2 \), where \( r_1 = \frac{\epsilon_0 h^2}{\pi m e^2} \) is the radius of the first Bohr orbit (ground state). This shows the radius depends on the square of the principal quantum number.
So, the ratio of radii for different orbits will be:
\( r_1 : r_2 : r_3 = (1)^2 : (2)^2 : (3)^2 : \dots \)
\( r_1 : r_2 : r_3 = 1 : 4 : 9 : \dots \)
In simple words: The size of an electron's orbit in a hydrogen atom grows much faster than its energy level number. If the first orbit has a certain radius, the second orbit will be four times bigger, and the third orbit will be nine times bigger, and so on.

🎯 Exam Tip: Remember that the radius of Bohr's orbit is directly proportional to the square of the principal quantum number (\( r_n \propto n^2 \)), leading to a rapid increase in size for higher orbits.

 

Question 7. What will be the electron's potential energy in eV in first orbit of hydrogen atom?
Answer: The total energy of an electron in the nth orbit of a hydrogen atom is given by \( E_n = -\frac{13.6}{n^2} \text{ eV} \).
The potential energy \( U_n \) of the electron in the nth orbit is twice its total energy:
\( U_n = 2 E_n \)
For the first orbit of a hydrogen atom, \( n=1 \).
First, calculate the total energy in the first orbit:
\( E_1 = -\frac{13.6}{(1)^2} = -13.6 \text{ eV} \)
Now, calculate the potential energy:
\( U_1 = 2 \times (-13.6 \text{ eV}) \)
\( U_1 = -27.2 \text{ eV} \)
In simple words: In the first orbit of a hydrogen atom, the electron's total energy is -13.6 eV. Its potential energy, which is twice the total energy, comes out to be -27.2 eV. The negative sign shows that the electron is bound to the nucleus.

🎯 Exam Tip: Always remember the relationship \( U_n = 2 E_n \) and \( K_n = -E_n \) for the potential and kinetic energies in a Bohr orbit. This relationship is derived from the electrostatic force and centripetal force balance.

 

Question 8. If radius of first Bohr orbit in hydrogen atom is 0.5 Å, then what will be radius in fourth Bohr orbit?
Answer: The radius of the \( n^{th} \) Bohr orbit is given by \( r_n = n^2 r_1 \), where \( r_1 \) is the radius of the first Bohr orbit.
Given: radius of the first Bohr orbit, \( r_1 = 0.5 \text{ Å} \).
We need to find the radius of the fourth Bohr orbit, so \( n = 4 \).
Using the formula:
\( r_4 = (4)^2 \times r_1 \)
\( r_4 = 16 \times 0.5 \text{ Å} \)
\( r_4 = 8.0 \text{ Å} \)
In simple words: The radius of an electron's orbit grows bigger by the square of the orbit number. So, if the first orbit is 0.5 Å wide, the fourth orbit will be 16 times wider, making it 8.0 Å.

🎯 Exam Tip: For problems involving Bohr orbit radii, recall the direct proportionality \( r_n \propto n^2 \) and simply multiply the first orbit's radius by the square of the desired orbit number.

 

Question 9. Calculate the wavelength of the last line of Balmer series.
Answer: The Rydberg formula for the Balmer series is:
\( \frac{1}{\lambda} = R \left[\frac{1}{2^2} - \frac{1}{n^2}\right] \)
For the last line (also known as the series limit) of the Balmer series, the electron transitions from an infinitely high energy level to \( n=2 \).
So, \( n = \infty \).
Substitute \( n = \infty \) into the formula:
\( \frac{1}{\lambda} = R \left[\frac{1}{2^2} - \frac{1}{\infty^2}\right] \)
Since \( \frac{1}{\infty^2} = 0 \):
\( \frac{1}{\lambda} = R \left[\frac{1}{4} - 0\right] \)
\( \frac{1}{\lambda} = \frac{R}{4} \)
Now, solve for \( \lambda \):
\( \lambda = \frac{4}{R} \)
Given Rydberg constant \( R = 1.097 \times 10^7 \text{ m}^{-1} \).
\( \lambda = \frac{4}{1.097 \times 10^7 \text{ m}^{-1}} \)
\( \lambda \approx 3.6463 \times 10^{-7} \text{ m} \)
To convert meters to angstroms (\( 1 \text{ m} = 10^{10} \text{ Å} \)):
\( \lambda = 3.6463 \times 10^{-7} \times 10^{10} \text{ Å} \)
\( \lambda = 3646.3 \text{ Å} \)
The source states \( \lambda = 3648 \text{ Å} \), which is a close approximation.
In simple words: The last line of the Balmer series is when an electron falls from an infinitely far distance to the second energy level. Using the Rydberg formula with the Rydberg constant, we calculate the wavelength of this light to be approximately 3648 Å.

🎯 Exam Tip: Remember that "last line" or "series limit" always refers to a transition from \( n = \infty \). Use the correct \( n_1 \) for the specific series (e.g., \( n_1=2 \) for Balmer).

 

Question 10. Write mathematical formulae for angular momentum related to quantisation in Bohr's theory.
Answer: According to Bohr's theory, the angular momentum (L) of an electron revolving in a stable orbit is quantized. This means that the angular momentum can only take specific, discrete values, not any continuous value. The mathematical formula for this quantization is:
\( L = n \frac{h}{2 \pi} \)
Where:
\( L \) = angular momentum of the electron
\( n \) = principal quantum number (an integer: 1, 2, 3, ...), which defines the orbit number
\( h \) = Planck's constant
\( \pi \) = pi (approximately 3.14)
The angular momentum can also be expressed as \( L = m v_n r_n \), where \( m \) is the mass of the electron, \( v_n \) is its velocity in the \( n^{th} \) orbit, and \( r_n \) is the radius of the \( n^{th} \) orbit. This equation is a fundamental principle that explains the stability of atoms and discrete energy levels.
In simple words: Bohr's theory says that electrons only orbit the nucleus in special paths where their spinning motion (angular momentum) is a whole number multiple of a basic tiny amount. This is written as \( L = n \frac{h}{2 \pi} \).

🎯 Exam Tip: Always state both the formula \( L = n \frac{h}{2\pi} \) and clearly define all the terms involved to score full marks on quantization of angular momentum.

 

Question 12. By which hypothesis Bohr's second postulate can be summarised?
Answer: Bohr's second postulate, which states that electrons can only revolve in orbits where their angular momentum is an integral multiple of \( \frac{h}{2\pi} \), can be explained or summarized by the de-Broglie matter wave hypothesis. This hypothesis suggests that electrons behave like waves, and only standing waves can exist in stable orbits, leading to quantized angular momentum. This wave-particle duality provides a deeper understanding of electron behavior in atoms.
In simple words: Bohr's idea that electrons only exist in specific orbits can be understood better by de-Broglie's idea that electrons also act like waves. Only waves that fit perfectly around the orbit without crashing into themselves are allowed.

🎯 Exam Tip: When connecting Bohr's postulates to modern physics, always highlight de-Broglie's hypothesis as the fundamental explanation for angular momentum quantization.

RBSE Class 12 Physics Chapter 14 Short Answer Type Questions

 

Question 1. Describe shortcomings of Thomson atomic model.
Answer: Thomson's atomic model, often called the "plum pudding" model, had several drawbacks that led to its eventual rejection:
1. It could not explain the origin of the various spectral series observed in the light emitted by atoms, such as hydrogen. It predicted a continuous spectrum instead of discrete lines.
2. It failed to explain the results of Rutherford's alpha-particle scattering experiment, which showed that a significant portion of alpha particles were deflected at large angles or even bounced back, indicating a concentrated positive charge within the atom.
Rutherford's experiment provided evidence against Thomson's model, which proposed a uniformly distributed positive charge.
In simple words: Thomson's model couldn't explain why atoms emit specific colors of light or why alpha particles bounced back in Rutherford's experiment, showing it wasn't quite right.

🎯 Exam Tip: Focus on the two major experimental observations that Thomson's model failed to explain: discrete spectra and large-angle alpha scattering.

 

Question 2. Describe Rutherford's model of the atom.
Answer: Rutherford's atomic model, based on his alpha-ray scattering experiment in 1911 (with Geiger and Marsden), proposed the following key features:
1. **Nucleus:** An atom has a very small, dense central core called the nucleus. This nucleus contains all the atom's positive charge and almost all its mass.
2. **Size of Nucleus:** The nucleus is extremely tiny (about \( 10^{-15} \) m in diameter) compared to the overall size of the atom (about \( 10^{-10} \) m), meaning most of the atom is empty space.
3. **Electrons and Charge Neutrality:** Electrons, which are negatively charged, revolve around the nucleus in various orbits, similar to planets orbiting the sun. The total negative charge of these electrons balances the total positive charge of the nucleus, making the atom electrically neutral overall.
4. **Electrostatic Force:** The centripetal force needed for the electrons to revolve around the nucleus is provided by the electrostatic attraction between the negatively charged electrons and the positively charged nucleus. This model represented a significant step towards understanding atomic structure.
In simple words: Rutherford said an atom has a tiny, heavy center called the nucleus with positive charge. Electrons spin around this nucleus like planets, and the empty space makes up most of the atom. The positive nucleus pulls the negative electrons to keep them in orbit.

🎯 Exam Tip: When describing Rutherford's model, ensure you cover the nucleus (its size, mass, charge), the orbiting electrons, and the role of electrostatic force in holding the atom together.

 

Question 3. Describe why Rutherford's model was not able to explain the stability of the atom.
Answer: Rutherford's model proposed that electrons orbit the nucleus like planets around the sun. However, classical electromagnetic theory states that any charged particle undergoing acceleration (like an orbiting electron changing direction) should continuously radiate energy. If an electron continuously radiates energy, it would lose energy and its orbit would gradually decrease in radius, causing it to spiral inwards and eventually collapse into the positively charged nucleus. This means the atom would not be stable. Since atoms are known to be stable, Rutherford's model could not explain this fundamental observation. This was a major limitation that Bohr's model later addressed by introducing quantization.
In simple words: Rutherford's model said electrons spin around the nucleus. But if they spin, they should lose energy and crash into the nucleus, making atoms unstable. Since atoms are stable, this was a big problem with his model.

🎯 Exam Tip: The instability of the atom predicted by Rutherford's model is a direct consequence of classical electromagnetic theory applied to an orbiting electron. Emphasize the continuous radiation of energy and the spiraling path.

 

Question 4. Write shortcomings of Bohr's theory.
Answer: Despite its successes, Bohr's theory had several limitations:
1. **Applicability:** Bohr's model was only successful in explaining the spectra of hydrogenic atoms (atoms with only one electron), such as \( \text{H}, \text{He}^+, \text{Li}^{++} \). It could not be applied to atoms with more than one electron because of the complex electron-electron interactions.
2. **Elliptical Orbits:** It did not explain why electrons should only choose circular orbits when elliptical orbits are also theoretically possible according to classical mechanics.
3. **Fine Structure:** The theory failed to explain the fine structure of spectral lines. High-resolution spectroscopes showed that what appeared to be single spectral lines were actually groups of very closely spaced lines, which Bohr's theory could not account for.
4. **Relative Intensities:** Bohr's theory could predict the frequencies of spectral lines but provided no information about their relative intensities. It could not tell which lines would be brighter or fainter.
5. **Zeeman and Stark Effects:** It could not explain the splitting of spectral lines when the atom is placed in a magnetic field (Zeeman effect) or an electric field (Stark effect).
6. **Electrical Forces:** The formulation of Bohr's model primarily focused on the electrical force between the nucleus and a single electron. It did not properly account for the electrical forces between multiple electrons in multi-electron atoms.
In simple words: Bohr's theory worked well for hydrogen but failed for other atoms, couldn't explain why electron orbits are only circular, and missed small details in light patterns like fine structure and magnetic field effects.

🎯 Exam Tip: When listing Bohr's shortcomings, classify them into areas like applicability (single electron), orbital geometry (circular vs. elliptical), spectral details (fine structure, intensities), and external field effects (Zeeman/Stark).

 

Question 6. How linear spectrum study is helpful in determination of elements?
Answer: The study of linear spectra is very helpful in determining and identifying different elements because each element has a unique set of energy levels for its electrons. When atoms of an element are excited, their electrons jump to higher energy levels and then fall back down, emitting light. The specific wavelengths (or frequencies) of this emitted light form a unique pattern of lines, known as a linear or atomic spectrum. Just like a fingerprint, this distinct pattern of spectral lines is characteristic of that particular element and no other. By analyzing the linear spectrum of an unknown sample and comparing it to known spectral patterns, scientists can accurately identify the elements present in the sample. This technique is widely used in astrophysics to identify elements in stars and galaxies, and in forensic science for material analysis. This unique characteristic makes spectroscopy an indispensable tool for elemental analysis.
In simple words: Every element has its own special "light fingerprint" made of specific colored lines. By looking at these unique light patterns, scientists can tell exactly which elements are present in a sample.

🎯 Exam Tip: Emphasize that each element possesses a unique electron energy level structure, which results in a distinct, characteristic line spectrum that acts as its "fingerprint."

 

Question 7. In a sample of hydrogen gas mostly atoms are in n = 1 energy level. On passing visible light through this gas some spectrum lines are absorbed. Which lines (Lyman or Balmer) spectrum are absorbed maximum and why?
Answer: When visible light is passed through a sample of hydrogen gas where most atoms are in the \( n=1 \) (ground state) energy level, the **Balmer series** lines will be absorbed most.
Here's why:
* **Lyman Series:** The Lyman series involves transitions to or from the \( n=1 \) energy level. Absorption lines for the Lyman series correspond to electrons jumping from \( n=1 \) to \( n=2, 3, 4, \dots \). However, these transitions require higher energies, which fall in the **ultraviolet** region of the electromagnetic spectrum, not visible light.
* **Balmer Series:** The Balmer series involves transitions to or from the \( n=2 \) energy level. Absorption lines for the Balmer series correspond to electrons jumping from \( n=2 \) to \( n=3, 4, 5, \dots \). For these transitions to occur, electrons must first be in the \( n=2 \) state. In a gas at normal temperatures, very few atoms are naturally in the \( n=2 \) excited state. However, the question states visible light is passed through, and asks which series is absorbed *maximum*. If the atoms absorb *visible* light, the transitions from \( n=2 \) (to \( n=3 \), \( n=4 \), etc.) correspond to the visible spectrum. For absorption to happen from the \( n=1 \) state in the visible range, the energy required would be much higher, falling outside the visible range.
* **Reconsideration:** The question asks which lines (Lyman or Balmer) are absorbed *maximum*. If most atoms are in \( n=1 \), then absorption would primarily be from \( n=1 \). Transitions from \( n=1 \) to higher states are the Lyman series. While these are typically UV, the question implies *some* absorption in visible light. However, the standard understanding is that if atoms are mostly in \( n=1 \), Lyman series (UV) would be absorbed. If it is *visible* light, then only atoms *already excited to n=2* would absorb it (Balmer). The provided answer on page 13 is "We get only Lyman series in absorption spectrum of hydrogen atom. Because in normal state all atoms are in lower state (n=1)." This directly contradicts the common understanding of *visible* light and Balmer series. Given the constraint to extract content *as written*, I will reproduce the source's answer that mentions Lyman series, and then add an enriching sentence and tip to clarify the typical context of visible light if possible without contradicting the source. * *Self-correction*: The source says "We get only Lyman series in absorption spectrum of hydrogen atom. Because in normal state all atoms are in lower state (n=1)." This implies absorption *from the ground state* (n=1), which correctly corresponds to the Lyman series. The "visible light" part of the question might be misleading or contextually off in the source. I must adhere to the source's answer. The absorption for Lyman series is in the UV range. Answer: In a sample of hydrogen gas where most atoms are in the \( n=1 \) energy level, the **Lyman series** lines will be absorbed maximum.
This is because the Lyman series corresponds to electron transitions from the ground state (\( n=1 \)) to higher energy levels (\( n=2, 3, 4, \dots \)). Since most of the atoms are already in the \( n=1 \) state (normal state), they are readily available to absorb photons that have the exact energy needed to make these transitions. While these transitions typically correspond to photons in the ultraviolet region of the electromagnetic spectrum, they represent the primary absorption events from the ground state. It is these specific energy jumps that characterize the Lyman series, making it the most absorbed series from the ground state. The energy levels are discrete, so only photons with exact energy differences will be absorbed.
In simple words: Since most hydrogen atoms are in their lowest energy state (\( n=1 \)), they will mostly absorb light that makes electrons jump from this \( n=1 \) level. These jumps are part of the Lyman series, which involves specific energy photons for absorption.

🎯 Exam Tip: For absorption, the electron must jump from its current energy level to a higher one. If atoms are primarily in the ground state (\( n=1 \)), then Lyman series (transitions from \( n=1 \)) will dominate absorption, typically in the UV range.

 

Question 8. What do you mean by stable orbit of the electron? What is the condition for it?
Answer: In Bohr's atomic model, a **stable orbit** (or stationary orbit) refers to a specific, permissible circular path around the nucleus in which an electron can revolve without radiating energy. According to classical physics, an orbiting electron should continuously lose energy and spiral into the nucleus. However, Bohr's model states that in these stable orbits, electrons do not emit or absorb energy, thus maintaining the atom's stability. These orbits are crucial for explaining why atoms don't collapse.
**Condition for Stable Orbits:**
The condition for an electron to exist in a stable orbit is that its angular momentum must be an integral multiple of \( \frac{h}{2\pi} \). Mathematically, this is expressed as:
\( L = n \frac{h}{2\pi} \)
Where:
\( L \) is the angular momentum of the electron.
\( n \) is the principal quantum number (an integer: 1, 2, 3, ...), representing the orbit number.
\( h \) is Planck's constant.
This quantization condition ensures that only certain discrete orbits are allowed, and electrons in these orbits are stable.
In simple words: Stable orbits are special paths where electrons circle the nucleus without losing energy, keeping the atom steady. The rule for these paths is that the electron's spin (angular momentum) must be a specific whole-number amount, like a step, not just any value.

🎯 Exam Tip: When defining stable orbits, clearly state the two key ideas: no energy radiation and the quantization condition for angular momentum (\( L = n \frac{h}{2\pi} \)).

 

Question 9. Balmer series was found firstly and analysed firstly than other lines. What is the cause of this reason?
Answer: The Balmer series was discovered and analyzed first among all the spectral series of hydrogen because some of its spectral lines fall within the **visible region** of the electromagnetic spectrum. This means that these lines could be easily observed and studied using early spectroscopic equipment, which primarily worked with visible light. In contrast, the Lyman series lies in the ultraviolet region, and the Paschen, Brackett, and Pfund series lie in the infrared region, all of which require more sophisticated detection equipment and were therefore discovered later. The visibility of Balmer lines made them readily accessible for early scientific investigation and analysis.
In simple words: The Balmer series was found first because some of its light lines are in colors we can see with our eyes. Other series produce light that is invisible, like ultraviolet or infrared, so they needed special tools to be found later.

🎯 Exam Tip: The key reason for the early discovery of the Balmer series is its visible light component. Always associate Balmer with visible light, Lyman with UV, and Paschen/Brackett/Pfund with IR.

RBSE Class 12 Physics Chapter 14 Long Answer Type Questions

 

Question 1. Describe Rutherford's experiment on scattering of a-particles. How do it help in discovery of nucleus?
Answer: Rutherford's alpha-ray scattering experiment, conducted by his colleagues Geiger and Marsden in 1911, was pivotal in discovering the atomic nucleus. The experimental setup involved a radioactive substance (like polonium) emitting alpha particles, placed in a lead box. These alpha particles were collimated into a narrow beam using diaphragms and directed onto a very thin gold foil. A fluorescent screen was placed around the gold foil to detect the scattered alpha particles; when an alpha particle hit the screen, it produced a tiny flash of light (scintillation) that could be observed through a microscope. The entire setup was kept in a vacuum to prevent scattering by air molecules.

**Observations and Conclusions:**
Rutherford observed the following:
(i) **Most alpha particles passed straight through the gold foil without any deflection.**
* **Conclusion:** This indicated that most of the atom is empty space. This observation contradicted Thomson's "plum pudding" model, which suggested a solid, uniformly positive atom.

(ii) **Some alpha particles were deflected from their path by small angles.**
* **Conclusion:** This suggested that the alpha particles (which are positively charged) encountered a positive charge within the atom. This positive charge must be concentrated, as a diffused charge (as in Thomson's model) would not cause such deflections.

(iii) **A very small number of alpha particles (about 1 in 8000) were deflected at very large angles (more than 90°) or even bounced back (180°).**
* **Conclusion:** This extremely rare but significant observation led Rutherford to conclude that all the positive charge and almost the entire mass of the atom are concentrated in an extremely tiny region at its center, which he called the **nucleus**. The strong repulsive force from this small, dense, positively charged nucleus was responsible for the backward scattering of alpha particles that approached it head-on. If the atom were uniformly positive, such large deflections would be impossible. The experiment essentially showed that the nucleus is a very small, dense, positively charged core.
This experiment demonstrated that the atom is not a uniform sphere but has a concentrated positive center, leading to the discovery of the atomic nucleus.
In simple words: Rutherford's experiment shot tiny particles at a thin gold sheet. Most went straight through, but a few bounced back or changed direction a lot. This showed that atoms have a tiny, heavy, positive center called the nucleus, with mostly empty space around it.

🎯 Exam Tip: When explaining Rutherford's experiment, ensure you describe the setup, each key observation (most pass through, some small deflections, very few large deflections/bounce back), and the corresponding conclusion for each observation, particularly the discovery of the tiny, dense, positive nucleus.

RBSE Class 12 Physics Chapter 14 Long Answer Type Questions

 

Question 1. Describe Rutherford's experiment on scattering of α-particles. How do it help in discovery of nucleus?
Answer:

Alpha Ray Scattering Experiment and Rutherford Model of Atom

In 1911, scientists Rutherford, Geiger, and Marsden conducted an experiment to understand the structure of an atom. They used a radioactive substance, polonium (or Radon), placed in a lead box. This substance emitted alpha particles, which are essentially fast-moving Helium nuclei (\( \text{2He}^4 \)) with high kinetic energy.

These alpha particles passed through diaphragms, creating a narrow beam. This beam then hit a very thin gold foil. The gold foil was chosen because it could be made extremely thin (about \( 10^{-5} \) cm) and its heavy nucleus was good for deflecting alpha particles. When alpha particles hit the foil, they deflected in various directions.

The deflection from the original path due to collisions with atoms in the gold foil is called scattering. The scattered alpha particles were observed using a fluorescent screen, which glowed when hit by an alpha particle. This allowed the scientists to count the particles at different angles using a movable microscope. The entire setup was kept in a vacuum to prevent air molecules from interfering.

Based on their observations, Rutherford drew several conclusions about the atom's structure:

(i) Most alpha particles passed straight through the gold foil without any deflection. This indicated that most of the atom is empty space. This finding contradicted Thomson's model, which proposed that an atom was a solid sphere of positive charge.

(ii) Some alpha particles were deflected at small angles. Since alpha particles are positively charged, their deflection could only be caused by another positive charge. This suggested that the total positive charge in an atom is concentrated in a very small region. This also went against Thomson's model, which stated that positive charge was spread evenly throughout the atom.

(iii) A very small number of alpha particles (about 1 in 8000) were deflected at angles of 90 degrees or more, or even bounced back. This showed that all of the atom's mass and positive charge are concentrated in a tiny central area, which Rutherford called the nucleus. This concentration of mass and charge in the nucleus is key to understanding atomic behavior.

The nucleus's existence successfully explained Rutherford's experimental results. Alpha particles far from the nucleus passed through without deflection. As an alpha particle got closer to the nucleus, the scattering angle increased due to the repulsive force.

Rutherford concluded that the repulsive force between the positive alpha particles and the positive nucleus follows Coulomb's law. This force causes the alpha particle's path to become hyperbolic. As the distance 'r' from the nucleus decreases, the repulsive force 'F' increases quickly. This means particles far away are hardly deflected, while those passing close to the nucleus are scattered at larger angles.

Rutherford calculated the number of alpha particles scattered at different angles, finding it inversely proportional to \( \text{sin}^4 (\theta / 2) \). This relationship was confirmed by Geiger and Marsden in 1993, proving that Coulomb's law applies even at atomic distances (down to \( 10^{-14} \) m).

In 1920, Chadwick discovered that the positive charge in the metal nucleus is \( Ze \), where \( Z \) is the atomic number and \( e \) is the electron's charge.

Distance of Closest Approach (Size of Nucleus): This is the minimum distance an energetic alpha particle, moving directly towards the nucleus, can reach before stopping and reversing its path. It is denoted by \( r_0 \).

Rutherford made these assumptions for calculating \( r_0 \):
(i) The atomic nucleus is so heavy that its motion during the impact is ignored.
(ii) Both the nucleus and the alpha particle are considered point charges with no dimensions.
(iii) The scattering is due to elastic collision between the nucleus and the alpha particle.

The kinetic energy of the alpha particle is transformed into potential energy as it approaches the nucleus. At the point of closest approach, all kinetic energy is converted into potential energy, and the particle briefly stops before turning back.

Kinetic energy of alpha particle: \( E_K = \frac{1}{2} mv^2 \)
Potential energy at closest approach: \( E_P = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_0} = \frac{1}{4 \pi \varepsilon_0} \frac{(+2e)(+Ze)}{r_0} = \frac{1}{4 \pi \varepsilon_0} \frac{2Ze^2}{r_0} \)
For stop motion, kinetic energy equals potential energy:
\( \frac{1}{2} mv^2 = \frac{1}{4 \pi \varepsilon_0} \frac{2Ze^2}{r_0} \)
Solving for \( r_0 \):
\( r_0 = \frac{1}{4 \pi \varepsilon_0} \frac{2Ze^2}{\frac{1}{2} mv^2} \)
\( r_0 = \frac{2Ze^2}{4 \pi \varepsilon_0 E_K} \)
This calculation helps determine the approximate size of the nucleus. The impact parameter is the perpendicular distance between the initial velocity vector of the alpha particle and the center of the nucleus. The scattering angle depends on this impact parameter.
In simple words: Rutherford shot tiny positive particles at a thin gold sheet. Most went straight through, but some bounced back or changed direction. This showed that atoms are mostly empty, but have a tiny, heavy, positive center called the nucleus. He used math to find how close the particles got to the nucleus, helping to estimate its size.

🎯 Exam Tip: When describing Rutherford's experiment, always include the key observations (most pass through, some deflect small angles, few deflect large angles) and their corresponding conclusions about the atom's structure (empty space, positive nucleus, dense nucleus).

 

Question 2. What are the shortcomings of Rutherford model. How are these shortcomings was removed by Bohr in his model?
Answer:

Drawbacks of Rutherford's Atomic Model
Rutherford's atomic model, despite its groundbreaking discovery of the nucleus, had several important limitations:

1. **Stability of Atom:** According to classical electromagnetic theory, an accelerated charged particle (like an electron orbiting the nucleus) must continuously emit electromagnetic radiation. If an electron continuously loses energy, it should spiral inwards and eventually fall into the nucleus. This would make the atom unstable, but atoms are generally stable. Rutherford's model could not explain why atoms do not collapse.
2. **Line Spectrum:** Rutherford's model predicted that electrons could revolve in any orbit, which would lead to a continuous spectrum of emitted radiation. However, atoms like hydrogen actually emit a discrete line spectrum, meaning they emit light at specific, distinct wavelengths. Rutherford's model could not explain this observation.
3. **Spectral Series Origin:** The model failed to explain the origin of various spectral series (like Lyman, Balmer, Paschen series) observed in the hydrogen atom.
4. **Electron Distribution:** It did not provide a clear picture of how electrons are distributed around the nucleus or their energy levels.

Bohr Model for Hydrogen Atom and Hydrogen-Like Ions:
Niels Bohr addressed these shortcomings in 1913 by proposing a new atomic model. He incorporated quantum theory (developed by Max Planck and Einstein) into Rutherford's model. Bohr's postulates were:

1. **Stationary Orbits:** Electrons can only revolve in certain specific, stable orbits around the nucleus without radiating energy. These are called "stationary orbits." An atom consists of a small, massive central nucleus, with electrons orbiting it. The electrostatic attraction between the electron and the nucleus provides the necessary centripetal force for the electron's revolution.
2. **Quantization of Angular Momentum:** Electrons are allowed to revolve only in those orbits where their angular momentum (\( L \)) is an integral multiple of \( \frac{h}{2\pi} \). Here, \( h \) is Planck's constant. This means \( L = n \frac{h}{2\pi} \), where \( n \) is a positive integer (principal quantum number). This condition selects specific, stable orbits.
3. **Energy Emission/Absorption:** An atom emits or absorbs radiation only when an electron jumps from one stationary orbit to another. If an electron moves from a higher energy orbit (\( E_{n2} \)) to a lower energy orbit (\( E_{n1} \)), it emits a photon. If it absorbs a photon, it moves from a lower to a higher energy orbit. The energy of the emitted or absorbed photon is given by \( \Delta E = E_{n2} - E_{n1} = hv = \frac{hc}{\lambda} \). This is known as Bohr's frequency condition.

Bohr's model successfully explained the stability of the atom and the discrete line spectra, especially for hydrogen. By introducing the concept of quantized energy levels and orbits, it resolved the major issues of Rutherford's model.
In simple words: Rutherford's model had two big problems: it said atoms should collapse (but they don't) and it couldn't explain why atoms emit only certain colors of light. Bohr fixed this by saying electrons only orbit in special paths without losing energy, and they only jump between these paths, emitting specific light colors when they do.

🎯 Exam Tip: Remember the two main shortcomings of Rutherford's model (atomic stability and continuous spectrum) and how Bohr's postulates (stationary orbits, quantized angular momentum, and specific energy transitions) directly addressed them.

 

Question 3. State Bohr's postulates for explaining the spectrum of hydrogen atoms. Derive a formula for total energy of electron in n orbit.
Answer:

Bohr's Postulates for Hydrogen Atom:

1. **Stationary Orbits:** An electron in an atom can revolve in certain non-radiating orbits, called stationary orbits, without losing energy. The centripetal force needed for this revolution is provided by the electrostatic attraction between the electron and the positively charged nucleus. So, \( F_{\text{coulomb}} = F_{\text{centripetal}} \).
   If an electron with charge \( e \) and mass \( m \) moves around a nucleus with charge \( Ze \) in an \( n^{\text{th}} \) stable orbit of radius \( r_n \) with velocity \( v_n \), then:
   \( \frac{k(Ze)e}{r_n^2} = \frac{mv_n^2}{r_n} \)
   Where \( k = \frac{1}{4\pi\epsilon_0} \).
2. **Quantization of Angular Momentum:** An electron can revolve only in those orbits for which its angular momentum is an integral multiple of \( \frac{h}{2\pi} \). Here, \( h \) is Planck's constant.
   So, angular momentum \( L = mv_n r_n = n \frac{h}{2\pi} \), where \( n = 1, 2, 3, ... \) is the principal quantum number. These are the stable orbits.
3. **Energy Emission and Absorption:** An atom emits or absorbs energy only when an electron makes a transition from a higher energy orbit to a lower energy orbit (emission) or from a lower to a higher energy orbit (absorption). The energy difference between the two orbits (\( E_{n2} - E_{n1} \)) is emitted or absorbed as a photon with energy \( hv \).
   \( E_{n2} - E_{n1} = hv = \frac{hc}{\lambda} \)
   This is known as Bohr's frequency condition.

Derivation of Total Energy of Electron in \( \text{n}^{\text{th}} \) Orbit:
From the first postulate, the centripetal force is provided by the electrostatic force:
\( \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_n^2} = \frac{mv_n^2}{r_n} \)
\( mv_n^2 = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_n} \) .... (Equation A)

From the second postulate, angular momentum is quantized:
\( mv_n r_n = n \frac{h}{2\pi} \)
\( v_n = \frac{nh}{2\pi m r_n} \) .... (Equation B)

Substitute \( v_n \) from (Equation B) into (Equation A):
\( m \left( \frac{nh}{2\pi m r_n} \right)^2 = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_n} \)
\( m \frac{n^2 h^2}{4\pi^2 m^2 r_n^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_n} \)
\( \frac{n^2 h^2}{4\pi^2 m r_n} = \frac{Ze^2}{4 \pi \varepsilon_0} \)
Now, solve for \( r_n \), the radius of the \( n^{\text{th}} \) orbit:
\( r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Ze^2} \) .... (Equation C)

Now, let's find the total energy (\( E_n \)) which is the sum of kinetic energy (\( K_n \)) and potential energy (\( U_n \)).
Kinetic energy: \( K_n = \frac{1}{2} mv_n^2 \)
From (Equation A), \( mv_n^2 = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_n} \)
So, \( K_n = \frac{1}{2} \left( \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_n} \right) = \frac{Ze^2}{8 \pi \varepsilon_0 r_n} \)

Potential energy of electron: This is the electrostatic potential energy between the nucleus (charge \( +Ze \)) and the electron (charge \( -e \)).
\( U_n = \frac{1}{4 \pi \varepsilon_0} \frac{(+Ze)(-e)}{r_n} = - \frac{Ze^2}{4 \pi \varepsilon_0 r_n} \)

Total energy: \( E_n = K_n + U_n \)
\( E_n = \frac{Ze^2}{8 \pi \varepsilon_0 r_n} - \frac{Ze^2}{4 \pi \varepsilon_0 r_n} \)
\( E_n = \frac{Ze^2}{8 \pi \varepsilon_0 r_n} - \frac{2Ze^2}{8 \pi \varepsilon_0 r_n} \)
\( E_n = - \frac{Ze^2}{8 \pi \varepsilon_0 r_n} \)

Now, substitute the value of \( r_n \) from (Equation C) into the expression for \( E_n \):
\( E_n = - \frac{Ze^2}{8 \pi \varepsilon_0 \left( \frac{n^2 h^2 \varepsilon_0}{\pi m Ze^2} \right)} \)
\( E_n = - \frac{Ze^2 \times \pi m Ze^2}{8 \pi \varepsilon_0 \times n^2 h^2 \varepsilon_0} \)
\( E_n = - \frac{Z^2 e^4 m}{8 \varepsilon_0^2 n^2 h^2} \)

This is the formula for the total energy of an electron in the \( n^{\text{th}} \) orbit of a hydrogen-like atom. For a hydrogen atom, \( Z=1 \), so the formula simplifies to:
\( E_n = - \frac{e^4 m}{8 \varepsilon_0^2 n^2 h^2} \)

Plugging in the constant values (\( e = 1.6 \times 10^{-19} \text{ C} \), \( m = 9.1 \times 10^{-31} \text{ kg} \), \( \varepsilon_0 = 8.86 \times 10^{-12} \text{ F/m} \), \( h = 6.63 \times 10^{-34} \text{ J s} \)), we get:
\( E_n = - \frac{13.6}{n^2} \text{ eV} \)
The negative sign indicates that the electron is bound to the nucleus and requires energy to be removed. As \( n \) increases, \( E_n \) becomes less negative (i.e., higher energy), approaching zero for very large \( n \).
In simple words: Bohr said electrons only exist in special orbits with fixed energy. They don't lose energy when in these orbits, but gain or lose it when jumping between them. We can use these ideas to calculate the total energy of an electron in any orbit by combining its movement energy and its stored energy due to its position near the nucleus. This calculation gives a negative energy value because the electron is trapped by the nucleus.

🎯 Exam Tip: Remember Bohr's three main postulates, especially the quantization of angular momentum. For the derivation, clearly show the steps for kinetic energy, potential energy, and then their sum, substituting the radius formula at the end to get the final energy expression.

 

Question 4. On the basis of Bohr atomic model explain the line spectrum of hydrogen atom.
Answer:
The Bohr atomic model successfully explains the line spectrum of hydrogen. A line spectrum is produced when light emitted from a gas or vapor, typically excited at low pressure by an electric current, contains only specific wavelengths. This is also known as an atomic spectrum.

Line Spectrum of Hydrogen and Its Explanation:
When atomic hydrogen gas is excited, its electrons absorb energy and jump from their normal (ground) state to higher energy levels (excited states). However, these excited states are unstable. The electrons quickly fall back to lower energy levels, including the ground state. During these transitions, electrons emit photons, and the energy of each photon corresponds exactly to the energy difference between the initial and final energy levels. Since these energy levels are discrete and quantized (as per Bohr's theory), only specific energy differences are possible. This results in the emission of photons with specific energies, and thus specific wavelengths, producing a line spectrum rather than a continuous one.

The hydrogen spectrum is characterized by several series of lines, named after their discoverers:

1. **Lyman Series:** Occurs when electrons transition from higher energy levels (\( n = 2, 3, 4, ... \)) to the ground state (\( n_1 = 1 \)). These lines are found in the ultraviolet region of the electromagnetic spectrum.
2. **Balmer Series:** Occurs when electrons transition from higher energy levels (\( n = 3, 4, 5, ... \)) to the second energy level (\( n_1 = 2 \)). Some lines of this series fall in the visible region, making them the first to be observed and studied, while others are in the near ultraviolet.
3. **Paschen Series:** Occurs when electrons transition from higher energy levels (\( n = 4, 5, 6, ... \)) to the third energy level (\( n_1 = 3 \)). These lines are found in the infrared region.
4. **Brackett Series:** Occurs when electrons transition from higher energy levels (\( n = 5, 6, 7, ... \)) to the fourth energy level (\( n_1 = 4 \)). These lines are also in the infrared region.
5. **Pfund Series:** Occurs when electrons transition from higher energy levels (\( n = 6, 7, 8, ... \)) to the fifth energy level (\( n_1 = 5 \)). These lines are in the far infrared region.

The wavelength \( \lambda \) for these series can be calculated using Rydberg's formula, derived from Bohr's model:
\( \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)
Where \( R \) is the Rydberg constant, \( n_1 \) is the principal quantum number of the lower energy level, and \( n_2 \) is the principal quantum number of the higher energy level (\( n_2 > n_1 \)).

Each element has a unique set of energy levels, resulting in a unique line spectrum. This makes the line spectrum a "fingerprint" for identifying elements. When white light passes through a gas, atoms absorb specific wavelengths, leading to an absorption line spectrum, which consists of dark lines on a continuous background. This is the reverse process of emission. The discrete nature of these spectral lines is direct evidence for the quantization of energy levels in atoms, a cornerstone of Bohr's model.
In simple words: Bohr's model explains why hydrogen gas, when excited, gives off light only in certain colors, not all colors. Electrons in atoms can only jump between specific energy levels. When they jump down, they release light of a precise color, creating distinct lines in the spectrum. Each type of jump creates a different "series" of lines, like the Lyman, Balmer, and Paschen series. This unique pattern of colors acts like a fingerprint for hydrogen.

🎯 Exam Tip: To explain the line spectrum, emphasize the discrete (quantized) energy levels and transitions between them. Mention at least the Lyman and Balmer series and their spectral regions. A clear explanation of how \( E = hv \) relates to \( \Delta E \) is crucial.

 

Question 5. Write shortcomings of Bohr's model. Show that Bohr's second postulate of orbital quantisation can be explained on the basis of De-Broglie hypothesis of wave nature of electron.
Answer:

Limitations of Bohr's Model:

Despite its success, Bohr's model had several limitations:

1. **Applicability to Hydrogenic Atoms Only:** The model works well only for hydrogenic atoms (atoms with a single electron), such as hydrogen, He\(^+ \), Li\(^{2+} \). It failed to explain the spectra of multi-electron atoms. The problem arises because electrons in multi-electron atoms interact with both the nucleus and each other, a complexity not addressed by Bohr's model.
2. **Circular Orbits:** Bohr's theory assumed electrons move in circular orbits. However, elliptical orbits are also possible, but the model did not explain why only circular orbits should be preferred.
3. **Fine Structure of Spectral Lines:** When observed with high-resolution spectroscopes, many spectral lines, especially in the hydrogen spectrum, are not single lines but a group of very closely spaced lines (fine structure). Bohr's theory could not explain these fine structures.
4. **Spectral Line Intensities:** The model could predict the frequencies of spectral lines but provided no information about their relative intensities (how bright each line is).
5. **Zeeman and Stark Effects:** Bohr's theory failed to explain the splitting of spectral lines into multiple components when the atom is placed in an external magnetic field (Zeeman effect) or an external electric field (Stark effect).
6. **Dual Nature of Electron:** Bohr's model treated electrons as particles moving in well-defined orbits, ignoring their wave nature.

Explanation of Bohr's Second Postulate by De-Broglie Hypothesis:

In 1927, Louis de Broglie proposed that particles like electrons also have wave-like properties. He suggested that the electron in its circular orbit, as described by Bohr, should be considered a particle wave. This idea provides a natural explanation for Bohr's second postulate (quantization of angular momentum).

Imagine a string tied at both ends. When plucked, only specific wavelengths (standing waves) can exist on the string. These standing waves have nodes at the ends and their length must be an integral multiple of half-wavelengths. For a wave to be stable and continuous in a circular orbit, it must form a standing wave pattern.

For an electron wave to form a stable standing wave in an orbit of radius \( r_n \), the circumference of the orbit must be an integral multiple of the electron's de Broglie wavelength \( \lambda \).
So, \( 2\pi r_n = n\lambda \), where \( n = 1, 2, 3, ... \)

According to de Broglie's hypothesis, the wavelength \( \lambda \) of a particle with momentum \( p \) is given by:
\( \lambda = \frac{h}{p} \)
If the electron's speed is much less than the speed of light, its momentum \( p \) can be written as \( mv_n \), where \( m \) is the electron's mass and \( v_n \) is its velocity in the \( n^{\text{th}} \) orbit.
So, \( \lambda = \frac{h}{mv_n} \)

Substitute this expression for \( \lambda \) into the standing wave condition:
\( 2\pi r_n = n \left( \frac{h}{mv_n} \right) \)
Rearranging this equation, we get:
\( mv_n r_n = n \frac{h}{2\pi} \)

This expression is exactly Bohr's second postulate, which states that the angular momentum of an electron in a stable orbit is an integral multiple of \( \frac{h}{2\pi} \). Therefore, the de Broglie hypothesis beautifully explains why only specific orbits are allowed for electrons—because only in these orbits can the electron waves form stable standing wave patterns without self-interference. This quantum condition, based on the wave nature of the electron, forms the basis for explaining the discrete orbits and energy levels in the hydrogen atom.
In simple words: Bohr's model couldn't explain why atoms have certain light patterns or why they stay stable. It also didn't work for atoms with many electrons. But later, de Broglie suggested that electrons act like waves. If an electron wave orbits the nucleus, it has to fit perfectly around the circle, like a vibrating guitar string. This means the orbit's size must be a whole number multiple of the electron's wavelength. When you write this math out, it becomes exactly Bohr's rule about angular momentum, showing that his rule came from the electron's wave nature.

🎯 Exam Tip: For shortcomings, focus on the inability to explain multi-electron atoms, fine structure, and Zeeman/Stark effects. For de Broglie's explanation, clearly state the standing wave condition (\( 2\pi r = n\lambda \)) and de Broglie's wavelength formula (\( \lambda = h/p \)), then show how they combine to yield Bohr's quantization condition.

 

Question 6. Derive a formula for radius of the stable orbit of hydrogen atom on the basis of Bohr model. Prove that in hydrogen atom the ratio of the stable radius are 1: 4: 9
Answer:

Derivation of Bohr's Radius or Radii of Stable Orbits:
Consider an atom with a nucleus of charge \( +Ze \) and an electron of mass \( m \) and charge \( -e \) revolving in the \( n^{\text{th}} \) orbit of radius \( r_n \) with velocity \( v_n \). The centripetal force required for the electron's revolution is provided by the electrostatic force of attraction between the electron and the nucleus.

According to Bohr's first postulate (Force balance):
\( \text{Electrostatic force} = \text{Centripetal force} \)
\( F_e = F_c \)
\( \frac{1}{4\pi\varepsilon_0} \frac{(Ze)(e)}{r_n^2} = \frac{mv_n^2}{r_n} \)
\( \frac{Ze^2}{4\pi\varepsilon_0 r_n^2} = \frac{mv_n^2}{r_n} \)
This simplifies to: \( mv_n^2 = \frac{Ze^2}{4\pi\varepsilon_0 r_n} \) .... (Equation 1)

According to Bohr's second postulate (Quantization of angular momentum):
The angular momentum of the electron in the \( n^{\text{th}} \) orbit is quantized:
\( L = mv_n r_n = n\frac{h}{2\pi} \)
From this, we can express \( v_n \):
\( v_n = \frac{nh}{2\pi m r_n} \) .... (Equation 2)

Now, substitute the expression for \( v_n \) from (Equation 2) into (Equation 1):
\( m \left( \frac{nh}{2\pi m r_n} \right)^2 = \frac{Ze^2}{4\pi\varepsilon_0 r_n} \)
\( m \frac{n^2 h^2}{4\pi^2 m^2 r_n^2} = \frac{Ze^2}{4\pi\varepsilon_0 r_n} \)
\( \frac{n^2 h^2}{4\pi^2 m r_n^2} = \frac{Ze^2}{4\pi\varepsilon_0 r_n} \)
We can cancel one \( r_n \) from both sides and rearrange to solve for \( r_n \):
\( \frac{n^2 h^2}{4\pi^2 m r_n} = \frac{Ze^2}{4\pi\varepsilon_0} \)
\( r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Ze^2} \)

This is the formula for the radius of the \( n^{\text{th}} \) stable orbit in a hydrogen-like atom. It shows that the radius is directly proportional to \( n^2 \).

Ratio of Stable Radii in Hydrogen Atom:
For a hydrogen atom, the atomic number \( Z = 1 \). So, the formula for the radius becomes:
\( r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \)

Let's define a constant \( a_0 = \frac{h^2 \varepsilon_0}{\pi m e^2} \), which is the Bohr radius (radius of the first orbit for hydrogen).
So, \( r_n = n^2 a_0 \).

For the first orbit (\( n = 1 \)): \( r_1 = (1)^2 a_0 = a_0 \)
For the second orbit (\( n = 2 \)): \( r_2 = (2)^2 a_0 = 4a_0 \)
For the third orbit (\( n = 3 \)): \( r_3 = (3)^2 a_0 = 9a_0 \)

Therefore, the ratio of the radii of the stable orbits in a hydrogen atom is:
\( r_1 : r_2 : r_3 = a_0 : 4a_0 : 9a_0 \)
\( r_1 : r_2 : r_3 = 1 : 4 : 9 \)

This proves that the radii of stable orbits are in the ratio of \( 1^2 : 2^2 : 3^2 \) for the first, second, and third orbits, respectively. The Bohr radius \( a_0 \) is approximately \( 0.529 \times 10^{-10} \text{ m} \) or \( 0.529 \text{ Å} \). This indicates that as the principal quantum number increases, the electron orbits become larger and further from the nucleus.
In simple words: To find the size of an electron's orbit, we balance the electrical pull from the nucleus with the force needed to keep the electron moving in a circle. We also use Bohr's rule that an electron's "spin" is fixed to certain values. Combining these rules, we get a formula for the orbit's radius. For a simple hydrogen atom, this formula shows that the radii of the first few orbits are in a simple ratio: 1 to 4 to 9, meaning the orbits get much bigger as you go further out.

🎯 Exam Tip: Clearly state and use Bohr's first two postulates to derive the radius formula. Emphasize that \( r_n \propto n^2 \) and then show the ratio derivation for \( n=1, 2, 3 \) explicitly. Remember that \( Z=1 \) for a hydrogen atom.

RBSE Class 12 Physics Chapter 14 Numerical Questions

 

Question 1. Calculate radius of second Bohr radius of hydrogen atom and then also calculate the speed of the electron and total energy of the orbit. (Given, mass of electron \( m = 9 \times 10^{-31} \text{ kg} \), \( e = 1.6 \times 10^{-19} \text{ C} \), \( h = 6.6 \times 10^{-34} \))
Answer:
Given values:
Mass of electron, \( m = 9.1 \times 10^{-31} \text{ kg} \)
Charge of electron, \( e = 1.6 \times 10^{-19} \text{ C} \)
Planck's constant, \( h = 6.6 \times 10^{-34} \text{ J s} \)
Coulomb's constant, \( k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2 \)
For hydrogen atom, atomic number \( Z=1 \). We need to calculate for the second orbit, so \( n=2 \).

1. Radius of the second Bohr orbit (\( r_2 \)):
The formula for the radius of the \( n^{\text{th}} \) orbit is:
\( r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Ze^2} \)
We know that \( a_0 = \frac{h^2 \varepsilon_0}{\pi m e^2} \) is the Bohr radius (approximately \( 0.529 \text{ Å} \)).
So, \( r_n = n^2 \frac{a_0}{Z} \)
For hydrogen \( Z=1 \), so \( r_n = n^2 a_0 \).
Given \( a_0 \approx 0.529 \text{ Å} \).
For \( n=2 \):
\( r_2 = (2)^2 \times 0.529 \text{ Å} \)
\( r_2 = 4 \times 0.529 \text{ Å} \)
\( r_2 = 2.116 \text{ Å} \)
This means the second orbit is four times larger than the first. An Ångström (\( \text{Å} \)) is a unit of length equal to \( 10^{-10} \) meters.

2. Speed of the electron in the second orbit (\( v_2 \)):
The formula for the velocity of an electron in the \( n^{\text{th}} \) orbit is:
\( v_n = \frac{Ze^2}{2\varepsilon_0 n h} \)
For hydrogen \( Z=1 \), and \( n=2 \):
\( v_2 = \frac{1 \times (1.6 \times 10^{-19})^2}{2 \times (8.854 \times 10^{-12}) \times 2 \times (6.63 \times 10^{-34})} \)
Alternatively, we use the known expression for velocity:
\( v_n = \frac{v_1}{n} Z \)
Where \( v_1 \) is the speed in the first Bohr orbit for hydrogen, \( v_1 \approx 2.18 \times 10^6 \text{ m/s} \).
For hydrogen (\( Z=1 \)) and \( n=2 \):
\( v_2 = \frac{2.18 \times 10^6 \text{ m/s}}{2} \times 1 \)
\( v_2 = 1.09 \times 10^6 \text{ m/s} \)
This velocity is significantly high, but much less than the speed of light, validating the non-relativistic approximation. The electron moves slower in higher energy orbits.

3. Total energy of the electron in the second orbit (\( E_2 \)):
The formula for the total energy of an electron in the \( n^{\text{th}} \) orbit is:
\( E_n = - \frac{Z^2 e^4 m}{8 \varepsilon_0^2 n^2 h^2} \)
For hydrogen \( Z=1 \):
\( E_n = - \frac{e^4 m}{8 \varepsilon_0^2 n^2 h^2} \)
We know that the ground state energy for hydrogen (\( Z=1, n=1 \)) is \( E_1 = -13.6 \text{ eV} \).
So, \( E_n = \frac{E_1}{n^2} \)
For \( n=2 \):
\( E_2 = \frac{-13.6 \text{ eV}}{(2)^2} \)
\( E_2 = \frac{-13.6 \text{ eV}}{4} \)
\( E_2 = -3.4 \text{ eV} \)
The energy in the second orbit is less negative than in the first, indicating a higher energy level. This means the electron is less tightly bound to the nucleus when it is in an orbit further away.
In simple words: For a hydrogen atom in its second orbit, we first calculate the orbit's size, which is about 2.116 Å. Then, we find the electron's speed in this orbit, which is about \( 1.09 \times 10^6 \text{ m/s} \). Finally, we calculate the total energy of the electron in this orbit, which comes out to be -3.4 eV, showing it's held by the nucleus but less tightly than in the first orbit.

🎯 Exam Tip: When calculating for hydrogen, remember that \( Z=1 \). Utilize the simplified formulas involving \( n^2 \) for radius and \( \frac{1}{n^2} \) for energy, as these are quick shortcuts from the ground state values. For velocity, remember it scales inversely with \( n \).

 

Question 2. If wavelength of the first line of Lyman series is 1216 Å, then find out wavelengths of the first lines of Balmer and Paschen series.
Answer:
For the Lyman series, the formula for wavelength \( \lambda_L \) is given by:
\( \frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{n_2^2} \right] \)
For the first line of the Lyman series, \( n_2 = 2 \). Substituting this value:
\( \frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] \)
\( \frac{1}{\lambda_L} = R \left[ 1 - \frac{1}{4} \right] \)
\( \frac{1}{\lambda_L} = R \left[ \frac{4-1}{4} \right] \)
\( \frac{1}{\lambda_L} = R \frac{3}{4} \)
Given \( \lambda_L = 1216 \) Å, so:
\( \frac{1}{1216} = \frac{3R}{4} \) ... (1)
For the Balmer series, the wavelength \( \lambda_B \) is calculated using the formula:
\( \frac{1}{\lambda_B} = R \left[ \frac{1}{2^2} - \frac{1}{n_2^2} \right] \)
For the first line of the Balmer series, \( n_2 = 3 \). Plugging this in:
\( \frac{1}{\lambda_B} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] \)
\( \frac{1}{\lambda_B} = R \left[ \frac{1}{4} - \frac{1}{9} \right] \)
\( \frac{1}{\lambda_B} = R \left[ \frac{9-4}{36} \right] \)
\( \frac{1}{\lambda_B} = R \frac{5}{36} \) ... (2)
Now, we divide equation (2) by equation (1) to find \( \lambda_B \):
\( \frac{1/\lambda_B}{1/1216} = \frac{5R/36}{3R/4} \)
\( \frac{1216}{\lambda_B} = \frac{5R}{36} \times \frac{4}{3R} \)
\( \frac{1216}{\lambda_B} = \frac{5 \times 4}{36 \times 3} \)
\( \frac{1216}{\lambda_B} = \frac{20}{108} \)
\( \lambda_B = 1216 \times \frac{108}{20} \)
\( \lambda_B = 1216 \times 5.4 \)
\( \lambda_B = 6566.4 \) Å
For the Paschen series, the wavelength \( \lambda_P \) is calculated using the formula:
\( \frac{1}{\lambda_P} = R \left[ \frac{1}{3^2} - \frac{1}{n_2^2} \right] \)
For the first line of the Paschen series, \( n_2 = 4 \). Substituting this value:
\( \frac{1}{\lambda_P} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] \)
\( \frac{1}{\lambda_P} = R \left[ \frac{1}{9} - \frac{1}{16} \right] \)
\( \frac{1}{\lambda_P} = R \left[ \frac{16-9}{144} \right] \)
\( \frac{1}{\lambda_P} = R \frac{7}{144} \) ... (3)
We divide equation (3) by equation (1) to find \( \lambda_P \):
\( \frac{1/\lambda_P}{1/1216} = \frac{7R/144}{3R/4} \)
\( \frac{1216}{\lambda_P} = \frac{7R}{144} \times \frac{4}{3R} \)
\( \frac{1216}{\lambda_P} = \frac{7 \times 4}{144 \times 3} \)
\( \frac{1216}{\lambda_P} = \frac{28}{432} \)
\( \lambda_P = 1216 \times \frac{432}{28} \)
\( \lambda_P = 1216 \times 15.428 \)
\( \lambda_P = 18764.5 \) Å
The Balmer series is in the visible region, while Lyman is ultraviolet and Paschen is infrared.
In simple words: We used the Rydberg formula to find the wavelengths for the first lines of the Balmer and Paschen series, given the wavelength of the first line of the Lyman series. This formula helps us predict the light emitted when electrons jump between energy levels in an atom.

🎯 Exam Tip: Remember the principal quantum numbers for the initial and final states for each series (Lyman n=1, Balmer n=2, Paschen n=3 for the final state) to correctly apply the Rydberg formula.

 

Question 3. In atom in transition from energy state A to C 1000 Å wavelength photons are emitted and in transition from B to C energy state 5000 Å wavelength photons are emitted. What will be the wavelength of photon in transition from A to B state?
Answer:
For the transition from energy state A to C, the energy difference \( E_A - E_C \) is related to the emitted photon's wavelength \( \lambda_1 \):
\( E_A - E_C = \frac{hc}{\lambda_1} \) ... (1)
Here, \( \lambda_1 = 1000 \) Å.
For the transition from energy state B to C, the energy difference \( E_B - E_C \) is related to the emitted photon's wavelength \( \lambda_2 \):
\( E_B - E_C = \frac{hc}{\lambda_2} \) ... (2)
Here, \( \lambda_2 = 5000 \) Å.
We want to find the wavelength \( \lambda_3 \) for the transition from A to B. This energy difference \( E_A - E_B \) can be found by subtracting the second equation from the first:
\( (E_A - E_C) - (E_B - E_C) = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} \)
\( E_A - E_B = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \)
Let \( E_A - E_B = \frac{hc}{\lambda_3} \). So:
\( \frac{hc}{\lambda_3} = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \)
Dividing both sides by \( hc \):
\( \frac{1}{\lambda_3} = \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \)
Combine the terms on the right side:
\( \frac{1}{\lambda_3} = \frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \)
This gives us the formula for \( \lambda_3 \):
\( \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_2 - \lambda_1} \)
Now, we plug in the given values:
\( \lambda_3 = \frac{1000 \times 5000}{5000 - 1000} \)
\( \lambda_3 = \frac{5000000}{4000} \)
\( \lambda_3 = 1250 \) Å
This calculation shows how energy conservation applies to photon emissions during atomic transitions.
In simple words: When an electron jumps between energy levels, it gives off light with a certain wavelength. If we know the wavelengths for jumps from A to C and B to C, we can calculate the wavelength for a jump directly from A to B by using the energy differences.

🎯 Exam Tip: Remember that energy differences are inversely proportional to wavelength. For transitions, energy is conserved, so you can combine equations for different transitions to find unknown wavelengths or energies.

 

Question 4. Doubly ionised Lithium atom whose atomic number is 3 is similar to hydrogen atom. (i) In this atom calculate wavelength of the radiation in excitation from first orbit to third orbit. (ii) How many spectrum lines will be observed in emission of excited system?
Answer:
(i) To calculate the wavelength of radiation for excitation from the first orbit (n=1) to the third orbit (n=3) in a doubly ionized Lithium atom (Li\(^{2+}\)), we use the energy formula for hydrogen-like atoms and the relation between energy and wavelength.
The energy of an electron in the n-th orbit of a hydrogen-like atom is given by:
\( E_n = -13.6 \frac{Z^2}{n^2} \) eV
For Lithium, the atomic number \( Z = 3 \).
Energy in the first orbit (n=1):
\( E_1 = -13.6 \frac{3^2}{1^2} = -13.6 \times 9 = -122.4 \) eV
Energy in the third orbit (n=3):
\( E_3 = -13.6 \frac{3^2}{3^2} = -13.6 \frac{9}{9} = -13.6 \) eV
The energy required for excitation (energy difference \( \Delta E \)) is:
\( \Delta E = E_3 - E_1 \)
\( \Delta E = -13.6 - (-122.4) \)
\( \Delta E = -13.6 + 122.4 \)
\( \Delta E = 108.8 \) eV
The wavelength \( \lambda \) of the radiation can be found using the formula \( \Delta E = \frac{hc}{\lambda} \). Given \( hc = 1242 \) eV-nm (a useful constant for such calculations):
\( \lambda = \frac{hc}{\Delta E} \)
\( \lambda = \frac{1242 \text{ eV-nm}}{108.8 \text{ eV}} \)
\( \lambda = 11.415 \) nm
Converting to Ångstroms (1 nm = 10 Å):
\( \lambda = 11.415 \times 10 = 114.15 \) Å
(ii) The number of spectral lines observed when an atom de-excites from an excited state 'n' to lower states is given by the formula:
Number of spectral lines \( = \frac{n(n-1)}{2} \)
Here, the atom is excited to the third orbit, so \( n = 3 \):
Number of spectral lines \( = \frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3 \)
The three possible transitions are from n=3 to n=2, n=3 to n=1, and n=2 to n=1. Each transition creates a unique spectral line.
In simple words: For a lithium ion that acts like hydrogen, we calculated the energy needed to jump from the first to the third energy level and the specific light wavelength it would absorb for this. Then, we found how many different kinds of light it could give off if it fell back down from the third level to lower ones.

🎯 Exam Tip: For hydrogen-like atoms, always use the atomic number Z in the energy formula. The number of emission lines formula helps predict all possible downward jumps from a given excited state, assuming all intermediate levels are allowed.

 

Question 5. The wavelength of first line of Balmer series is 6564 Å, then find Rydberg constant and wave number.
Answer:
We are given that the wavelength of the first line of the Balmer series is \( \lambda = 6564 \) Å. We need to find the Rydberg constant (R) and the wave number \( \bar{v} \).
The wave number \( \bar{v} \) is the reciprocal of the wavelength:
\( \bar{v} = \frac{1}{\lambda} \)
First, convert the wavelength from Ångstroms to meters: \( 6564 \) Å \( = 6564 \times 10^{-10} \) m.
\( \bar{v} = \frac{1}{6564 \times 10^{-10} \text{ m}} \)
\( \bar{v} \approx 1.523 \times 10^6 \text{ m}^{-1} \)
For the Balmer series, the formula for wavelength is:
\( \frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n_2^2} \right] \)
For the first line of the Balmer series, the electron transitions from \( n_2 = 3 \) to \( n_1 = 2 \). So, we plug in \( n_2 = 3 \):
\( \frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] \)
\( \frac{1}{6564 \times 10^{-10}} = R \left[ \frac{1}{4} - \frac{1}{9} \right] \)
\( \frac{1}{6564 \times 10^{-10}} = R \left[ \frac{9-4}{36} \right] \)
\( \frac{1}{6564 \times 10^{-10}} = R \frac{5}{36} \)
Now, we can solve for R:
\( R = \frac{36}{5 \times 6564 \times 10^{-10}} \)
\( R = \frac{36}{32820 \times 10^{-10}} \)
\( R = \frac{36000 \times 10^7}{32820} \)
\( R \approx 1.0968 \times 10^7 \text{ m}^{-1} \)
So, the Rydberg constant is approximately \( 1.097 \times 10^7 \text{ m}^{-1} \). This constant is fundamental in atomic physics for predicting spectral lines.
In simple words: We used the given wavelength for a specific light line in the Balmer series to find two important values. First, the wave number, which tells us how many waves fit into a certain length. Second, the Rydberg constant, which is a key number that helps calculate the wavelengths of light emitted by hydrogen atoms.

🎯 Exam Tip: Remember to convert wavelength units (Å to m) consistently before calculations. Also, correctly identify \(n_1\) and \(n_2\) for the specific spectral series and line (e.g., first line means \(n_2 = n_1 + 1\)).

 

Question 6. Hydrogen similar any ion emits a radiation of 2.467 × 10⁷ Hz frequency in transition from n = 2 to n = 1. Calculate the frequency of the radiation in transition from n = 3 to n = 1.
Answer:
We know the frequency \( \nu \) of emitted radiation is related to the Rydberg constant R and the speed of light c by:
\( \nu = c R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \)
Let's analyze the first transition from \( n_2 = 2 \) to \( n_1 = 1 \). The given frequency is \( \nu_1 = 2.467 \times 10^7 \) Hz.
\( \nu_1 = c R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] \)
\( \nu_1 = c R \left[ 1 - \frac{1}{4} \right] \)
\( \nu_1 = c R \left[ \frac{3}{4} \right] \) ... (1)
Now, we need to calculate the frequency \( \nu_2 \) for the transition from \( n_2 = 3 \) to \( n_1 = 1 \):
\( \nu_2 = c R \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] \)
\( \nu_2 = c R \left[ 1 - \frac{1}{9} \right] \)
\( \nu_2 = c R \left[ \frac{8}{9} \right] \) ... (2)
To find \( \nu_2 \), we can divide equation (2) by equation (1):
\( \frac{\nu_2}{\nu_1} = \frac{c R (8/9)}{c R (3/4)} \)
\( \frac{\nu_2}{\nu_1} = \frac{8/9}{3/4} \)
\( \frac{\nu_2}{\nu_1} = \frac{8}{9} \times \frac{4}{3} \)
\( \frac{\nu_2}{\nu_1} = \frac{32}{27} \)
Now, solve for \( \nu_2 \):
\( \nu_2 = \nu_1 \times \frac{32}{27} \)
\( \nu_2 = 2.467 \times 10^7 \text{ Hz} \times \frac{32}{27} \)
\( \nu_2 = 2.467 \times 10^7 \times 1.185 \)
\( \nu_2 \approx 2.922 \times 10^7 \) Hz
This calculation demonstrates how energy transitions correspond to specific light frequencies.
In simple words: We used the given frequency of light emitted when an electron in a hydrogen-like ion jumps from level 2 to level 1. Then, we used this information and the general formula for frequency in atomic transitions to calculate the new frequency when an electron jumps from level 3 to level 1.

🎯 Exam Tip: When dealing with transitions in hydrogen-like atoms, recognize that the product 'cR' is constant for a given ion, allowing you to find ratios of frequencies or wavelengths for different transitions.

 

Question 7. Monochromatic radiation of wavelength λ are incident on a hydrogen sample whose atoms are in ground state. Hydrogen atom absorbs radiation and then emit wavelengths of 6 different colours. Determine λ. (Given he = 1242 eV-nm, ground state energy of hydrogen E = -13.6 eV)
Answer:
A hydrogen atom in its ground state (\(n=1\)) absorbs radiation. If it then emits 6 different wavelengths (colours), this implies that the electron was excited to a higher energy level and then de-excited through multiple steps. The number of spectral lines (N) emitted from an excited state 'n' is given by:
\( N = \frac{n(n-1)}{2} \)
We are given that \( N = 6 \). We need to find the excited state 'n':
\( 6 = \frac{n(n-1)}{2} \)
\( 12 = n(n-1) \)
We can test small integer values for 'n':
If \( n = 1 \), \( 1(0) = 0 \)
If \( n = 2 \), \( 2(1) = 2 \)
If \( n = 3 \), \( 3(2) = 6 \)
If \( n = 4 \), \( 4(3) = 12 \)
So, the atom was excited to the \( n=4 \) energy level.
Now, we need to find the wavelength \( \lambda \) of the incident radiation that caused this excitation from the ground state (\( n=1 \)) to \( n=4 \).
The energy of an electron in the n-th orbit of a hydrogen atom is:
\( E_n = -13.6 \frac{1^2}{n^2} \) eV (since Z=1 for hydrogen)
Ground state energy (\( n=1 \)):
\( E_1 = -13.6 \frac{1}{1^2} = -13.6 \) eV
Excited state energy (\( n=4 \)):
\( E_4 = -13.6 \frac{1}{4^2} = -13.6 \frac{1}{16} = -0.85 \) eV
The energy absorbed (\( \Delta E \)) for this transition is:
\( \Delta E = E_4 - E_1 \)
\( \Delta E = -0.85 - (-13.6) \)
\( \Delta E = -0.85 + 13.6 \)
\( \Delta E = 12.75 \) eV
The wavelength \( \lambda \) of the incident radiation is related to \( \Delta E \) by \( \Delta E = \frac{hc}{\lambda} \). Given \( hc = 1242 \) eV-nm:
\( \lambda = \frac{hc}{\Delta E} \)
\( \lambda = \frac{1242 \text{ eV-nm}}{12.75 \text{ eV}} \)
\( \lambda = 97.41 \) nm
This wavelength corresponds to the energy needed to excite the hydrogen atom to the \( n=4 \) state.
In simple words: A hydrogen atom in its lowest energy state absorbed light and then gave off 6 different kinds of light. We used this information to figure out that the atom was excited to the fourth energy level. Then, we calculated the exact wavelength of the light that was absorbed to make it jump to that level.

🎯 Exam Tip: Always relate the number of emitted spectral lines to the highest excited state 'n' using the formula \(N = n(n-1)/2\). This 'n' is crucial for calculating energy differences and wavelengths for the absorption process.

 

Question 8. In hydrogen atom light corresponding to transition from n = 4 to n = 2 falls on a metal whose work function is 1.9 eV. Find out the maximum kinetic energy of the emitted photoelectrons.
Answer:
First, we need to find the energy of the photon emitted during the transition from \(n=4\) to \(n=2\) in a hydrogen atom. This energy will be the energy of the incident light on the metal.
The energy of an electron in the n-th orbit of a hydrogen atom is given by:
\( E_n = -13.6 \frac{1^2}{n^2} \) eV (since Z=1 for hydrogen)
Energy in the \( n=4 \) orbit:
\( E_4 = -13.6 \frac{1}{4^2} = -13.6 \frac{1}{16} = -0.85 \) eV
Energy in the \( n=2 \) orbit:
\( E_2 = -13.6 \frac{1}{2^2} = -13.6 \frac{1}{4} = -3.4 \) eV
The energy of the emitted photon (\( \Delta E \)) is the difference between these energy levels:
\( \Delta E = E_4 - E_2 \)
\( \Delta E = -0.85 - (-3.4) \)
\( \Delta E = -0.85 + 3.4 \)
\( \Delta E = 2.55 \) eV
This \( \Delta E \) is the energy of the incident photon (\(hv\)) on the metal surface.
Now, we use Einstein's photoelectric equation to find the maximum kinetic energy (\(E_K\)_{max}) of the emitted photoelectrons:
\( hv = W_0 + (E_K)_{\text{max}} \)
Here, \( W_0 \) is the work function of the metal, given as \( 1.9 \) eV.
\( (E_K)_{\text{max}} = hv - W_0 \)
\( (E_K)_{\text{max}} = 2.55 \text{ eV} - 1.9 \text{ eV} \)
\( (E_K)_{\text{max}} = 0.65 \) eV
This calculation shows the energy conversion from photon to electron kinetic energy.
In simple words: When an electron in a hydrogen atom jumps from a higher energy level (n=4) to a lower one (n=2), it releases a photon of light. This light then hits a metal surface. We found out how much energy this light carries. Since we also know how much energy is needed to just get an electron out of the metal (work function), we subtracted that from the photon's energy to find the maximum leftover energy that the emitted electron has as kinetic energy.

🎯 Exam Tip: For photoelectric effect problems, first calculate the incident photon energy (often from energy level transitions) and then apply Einstein's equation: \(hv = W_0 + (E_K)_{\text{max}}\).

 

Question 9. A sample of hydrogen is in a specific excited state A. By absorption of photons of energy 2.55 eV it reaches to another excited state B. Calculate principal quantum number for A and B. [NA = 2, NB = 4]
Answer:
We are given that a hydrogen atom in excited state A absorbs a photon of energy \( \Delta E = 2.55 \) eV and reaches excited state B. We need to find the principal quantum numbers for states A (\(n_A\)) and B (\(n_B\)).
The energy of an electron in the n-th orbit of a hydrogen atom is:
\( E_n = -13.6 \frac{1}{n^2} \) eV (since Z=1 for hydrogen)
The energy absorbed is the difference between the final and initial energy states:
\( \Delta E = E_B - E_A \)
\( 2.55 \text{ eV} = \left( -13.6 \frac{1}{n_B^2} \right) - \left( -13.6 \frac{1}{n_A^2} \right) \)
\( 2.55 = 13.6 \left( \frac{1}{n_A^2} - \frac{1}{n_B^2} \right) \)
Dividing by 13.6:
\( \frac{2.55}{13.6} = \frac{1}{n_A^2} - \frac{1}{n_B^2} \)
\( 0.1875 = \frac{1}{n_A^2} - \frac{1}{n_B^2} \)
We need to find two integer values \( n_A \) and \( n_B \) (where \( n_B > n_A \)) that satisfy this equation.
Let's test common energy levels for hydrogen:
If \( n_A = 1 \), \( E_1 = -13.6 \) eV
If \( n_A = 2 \), \( E_2 = -13.6/4 = -3.4 \) eV
If \( n_A = 3 \), \( E_3 = -13.6/9 = -1.51 \) eV
If \( n_A = 4 \), \( E_4 = -13.6/16 = -0.85 \) eV
Let's try \( n_A = 2 \). Then \( E_A = -3.4 \) eV.
So, \( E_B = E_A + \Delta E = -3.4 \text{ eV} + 2.55 \text{ eV} = -0.85 \) eV.
Now, we need to find \( n_B \) such that \( E_{n_B} = -0.85 \) eV.
We know \( E_4 = -0.85 \) eV.
Therefore, \( n_B = 4 \).
So, the principal quantum number for state A is \( n_A = 2 \) and for state B is \( n_B = 4 \). This transition corresponds to the second line of the Paschen series.
In simple words: A hydrogen atom in some excited state absorbed energy and moved to an even higher excited state. We used the amount of energy absorbed and the formula for energy levels in a hydrogen atom to figure out which specific energy levels (A and B) the electron jumped between.

🎯 Exam Tip: When given an energy difference, it's often easiest to calculate the actual energy levels (\(E_n\)) for a few common 'n' values and then test which pair of levels matches the given energy difference. This helps in quickly identifying \(n_A\) and \(n_B\).

 

Question 10. The energy state diagram of an atom is shown in figure given below. Calculate wavelength of photons corresponding to transition B and D.
Answer:
(Note: The energy state diagram is not provided in the OCR text. We will proceed with the calculation assuming the energy values for states B and D are available from the diagram as implied by the solution's first line: \(E_{n_2} - E_{n_1}\) for D.)
For transition B, the energy difference is given as \( \Delta E = 0 - (-4.5) = 4.5 \) eV.
The wavelength \( \lambda \) of the photon is related to its energy \( \Delta E \) by \( \lambda = \frac{hc}{\Delta E} \). Using \( hc = 1242 \) eV-nm (a standard constant):
\( \lambda_B = \frac{1242 \text{ eV-nm}}{4.5 \text{ eV}} \)
\( \lambda_B = 276 \) nm
Converting to Ångstroms (1 nm = 10 Å):
\( \lambda_B = 276 \times 10 = 2760 \) Å
For transition D, the energy difference is given as \( \Delta E = -2 - (-10) = 8 \) eV.
The wavelength \( \lambda_D \) of the photon is:
\( \lambda_D = \frac{hc}{\Delta E} \)
\( \lambda_D = \frac{1242 \text{ eV-nm}}{8 \text{ eV}} \)
\( \lambda_D = 155.25 \) nm
Converting to Ångstroms:
\( \lambda_D = 155.25 \times 10 = 1552.5 \) Å
These calculations link discrete energy transitions to specific wavelengths of light emitted.
In simple words: Given a diagram of atom's energy levels, we found the energy difference for two specific electron jumps (transitions B and D). Then, using a special constant, we calculated the exact wavelengths of light photons that would be released during each of these jumps.

🎯 Exam Tip: Always clearly identify the initial and final energy levels for each transition from the given energy diagram. Use the constant \(hc\) (Planck's constant times speed of light) in convenient units (e.g., eV-nm) to quickly convert between energy and wavelength.

 

Question 11. Calculate maximum angular velocity of the electron in stable orbit of a hydrogen atom.
Answer:
The angular velocity \( \omega \) of an electron in an orbit is given by \( \omega = \frac{v}{r} \), where \( v \) is the linear velocity and \( r \) is the radius of the orbit.
For a hydrogen atom (Z=1), the radius of the n-th Bohr orbit is:
\( r_n = n^2 r_1 \)
where \( r_1 = 0.529 \times 10^{-10} \) m (Bohr radius). So, \( r_n = 0.529 n^2 \times 10^{-10} \) m.
The linear velocity of the electron in the n-th orbit for a hydrogen-like atom (Z=1 for hydrogen) is:
\( v_n = \frac{Z v_1}{n} = \frac{v_1}{n} \)
where \( v_1 = 2.18 \times 10^6 \) m/s (velocity in the first Bohr orbit). So, \( v_n = \frac{2.18 \times 10^6}{n} \) m/s.
Now, we can find the angular velocity \( \omega_n \):
\( \omega_n = \frac{v_n}{r_n} = \frac{(2.18 \times 10^6 / n)}{(0.529 n^2 \times 10^{-10})} \)
\( \omega_n = \frac{2.18 \times 10^6}{0.529 n^3 \times 10^{-10}} \)
\( \omega_n = \frac{2.18 \times 10^{16}}{0.529 n^3} \)
\( \omega_n \approx \frac{4.12 \times 10^{16}}{n^3} \) rad/s
To find the maximum angular velocity, we need the smallest value for \( n \), which is \( n=1 \) (ground state).
For \( n=1 \):
\( \omega_1 = \frac{4.12 \times 10^{16}}{1^3} \)
\( \omega_1 = 4.12 \times 10^{16} \) rad/s
This angular velocity is extremely high, reflecting the electron's rapid motion around the nucleus. The angular velocity decreases rapidly with increasing principal quantum number.
In simple words: We calculated how fast an electron spins around the center of a hydrogen atom. We used formulas for the electron's speed and the size of its path. The fastest spin happens when the electron is in its innermost path (ground state, n=1).

🎯 Exam Tip: Remember that angular velocity for an electron in a Bohr orbit is inversely proportional to \(n^3\). Therefore, the maximum angular velocity occurs for the smallest principal quantum number, \(n=1\).

 

Question 12. What will be the recoiling momentum of hydrogen atom in going from n = 5 state to n = 1 state after emission of photon? (Given, R = 1.097 x 10⁷ m⁻¹, h = 6.63 × 10⁻³⁴ J-s and mass of hydrogen = 1.67 x 10⁻²⁷ kg).
Answer:
When a photon is emitted, the atom recoils to conserve momentum. The recoiling momentum of the hydrogen atom will be equal in magnitude and opposite in direction to the momentum of the emitted photon.
The momentum of a photon \( p \) is given by \( p = \frac{\Delta E}{c} \), where \( \Delta E \) is the energy of the photon and \( c \) is the speed of light.
First, we calculate the energy of the photon emitted during the transition from \( n=5 \) to \( n=1 \) in a hydrogen atom.
The energy of an electron in the n-th orbit of a hydrogen atom is:
\( E_n = -13.6 \frac{1}{n^2} \) eV (since Z=1 for hydrogen)
Energy in the \( n=5 \) orbit:
\( E_5 = -13.6 \frac{1}{5^2} = -13.6 \frac{1}{25} = -0.544 \) eV
Energy in the \( n=1 \) orbit (ground state):
\( E_1 = -13.6 \frac{1}{1^2} = -13.6 \) eV
The energy of the emitted photon \( \Delta E \) is:
\( \Delta E = E_5 - E_1 \)
\( \Delta E = -0.544 - (-13.6) \)
\( \Delta E = -0.544 + 13.6 \)
\( \Delta E = 13.056 \) eV
Now, we need to convert this energy from eV to Joules (1 eV = \( 1.6 \times 10^{-19} \) J):
\( \Delta E = 13.056 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} \)
\( \Delta E = 20.8896 \times 10^{-19} \) J
\( \Delta E \approx 20.89 \times 10^{-19} \) J
Now we can calculate the momentum of the photon. The speed of light \( c \approx 3 \times 10^8 \) m/s.
\( p = \frac{\Delta E}{c} \)
\( p = \frac{20.89 \times 10^{-19} \text{ J}}{3 \times 10^8 \text{ m/s}} \)
\( p \approx 6.963 \times 10^{-27} \) kg m/s
Therefore, the recoiling momentum of the hydrogen atom is \( 6.963 \times 10^{-27} \) kg m/s. This effect, though small, is an important demonstration of momentum conservation at the atomic level.
In simple words: When a hydrogen atom's electron jumps from a higher energy level (n=5) to a lower one (n=1), it releases a tiny packet of light called a photon. Because of a rule called conservation of momentum, the atom itself gets a small "kick" in the opposite direction, like a gun recoiling when it shoots a bullet. We calculated the strength of this kick by finding the energy of the photon and then its momentum.

🎯 Exam Tip: Remember to convert the photon energy from eV to Joules before using it to calculate momentum in kg m/s. Also, recognize that the atom's recoil momentum is equal in magnitude to the photon's momentum.

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