Samacheer Kalvi Class 12 Physics Solutions Chapter 7 Dual Nature of Radiation and Matter

🎯 Exam Tip: Remember that de Broglie wavelength is inversely proportional to the square root of the accelerating potential, meaning higher voltage results in a shorter wavelength.

Question 12. Briefly explain the principle and working of an electron microscope.
Answer: An electron microscope utilizes the wave nature of electrons to magnify tiny objects, allowing us to see details much smaller than what is possible with visible light. It works on the principle that the resolving power of a microscope is inversely proportional to the wavelength of the radiation used. This means shorter wavelengths provide better resolution. Since electrons have a much shorter de Broglie wavelength compared to visible light, electron microscopes offer significantly higher resolving power and magnification.
For example, electron microscopes can magnify objects over 200,000 times, which is far greater than optical microscopes.
**Working:**
1. The overall design and function of an electron microscope are similar to an optical microscope, but it uses electrons instead of light.
2. Instead of glass lenses, electron microscopes use electrostatic or magnetic lenses to focus the electron beam.
3. The divergence and convergence of the electron beam can be controlled by adjusting these electric and magnetic fields.
4. Electrons are accelerated from a source by a high potential, and then these electrons are made parallel by a magnetic condenser lens.
5. A magnetic objective lens and a magnetic projector lens system help to form and magnify the image.
6. These microscopes are widely used in various scientific fields because of their superior resolution.
In simple words: An electron microscope uses tiny electron waves instead of light waves to make very small things look much bigger. Because electron waves are much smaller than light waves, these microscopes can see much finer details than regular microscopes.

🎯 Exam Tip: Highlight that the key advantage of an electron microscope is its extremely high resolving power due to the very short de Broglie wavelength of electrons.

Question 13. Describe briefly Davisson – Germer experiment which demonstrated the wave nature of electrons.
Answer: The Davisson-Germer experiment provided direct experimental proof for the wave nature of electrons, confirming de Broglie's hypothesis of matter waves. This groundbreaking experiment was conducted by Clinton Davisson and Lester Germer in 1927.
**Observations:**
* They showed that electron beams behave like waves and can be diffracted. This happens when electron beams strike the regular arrangement of atoms in crystalline solids.
* Since crystals have a structured, repeating arrangement of atoms, they can act like a three-dimensional diffraction grating for matter waves.
* When electron waves hit these crystals, they scatter in specific directions, creating diffraction patterns. This is similar to how light waves are diffracted by a grating.
* In their setup, a filament \( F \) was heated by a low tension battery, causing electrons to be emitted through thermionic emission.
* These electrons were then accelerated by a potential difference between the filament and an anode aluminum cylinder, powered by a high tension battery. This created a collimated electron beam.
* This electron beam was directed to strike a single crystal of nickel, which acted as the diffraction grating, using two thin aluminum diaphragms to focus the beam.
* An electron detector was used to measure the intensity of the scattered electron beam at various angles \( \theta \).
* They found that for an accelerating voltage of 54 V, the scattered electron wave showed a peak intensity at an angle of 50° to the incident beam, which is a characteristic of diffraction.
* This observed constructive interference of diffracted electrons perfectly matched the predictions for the interplanar spacing of Nickel.
* The de-Broglie wavelength for \( V = 54 \text{ V} \) was calculated as:
\( \lambda = \frac{12.27}{\sqrt{\mathrm{V}}} \text{ Å} \)
\( \lambda = \frac{12.27}{\sqrt{54}} \)
\( \lambda \approx 1.67 \text{ Å} \)
This confirmed that electrons indeed exhibit wave-like properties, a fundamental concept in quantum mechanics.
In simple words: The Davisson-Germer experiment proved that tiny electrons act like waves, not just particles. They found that when electrons hit a crystal, they bounce off in a pattern, just like light waves would. This showed that everything, even tiny particles, can have wave-like properties.

🎯 Exam Tip: When describing the Davisson-Germer experiment, emphasize that it was the first direct experimental evidence for the wave nature of matter, supporting de Broglie's hypothesis and forming a cornerstone of quantum physics.

IV. Numerical Problems:

Question 1. How many photons per second emanate from a 50 mW laser of 640 nm?
Answer:
Given:
Power \( P = 50 \times 10^{-3} \text{ W} \)
Wavelength \( \lambda = 640 \times 10^{-9} \text{ m} \)
We need to find \( N \), the number of photons emitted per second.
The power \( P \) can be expressed as the number of photons \( N \) multiplied by the energy of a single photon \( hv \). The energy \( hv \) can also be written as \( \frac{hc}{\lambda} \).
So, \( P = N \frac{hc}{\lambda} \)
Rearranging the equation to solve for \( N \):
\( N = \frac{P \lambda}{hc} \)
Now, we plug in the given values and the constants (Planck's constant \( h = 6.625 \times 10^{-34} \text{ Js} \) and speed of light \( c = 3 \times 10^8 \text{ m/s} \)):
\( N = \frac{50 \times 10^{-3} \times 640 \times 10^{-9}}{6.625 \times 10^{-34} \times 3 \times 10^8} \)
First, calculate the numerator:
\( 50 \times 10^{-3} \times 640 \times 10^{-9} = 32000 \times 10^{-12} = 3.2 \times 10^{-8} \)
Next, calculate the denominator:
\( 6.625 \times 10^{-34} \times 3 \times 10^8 = 19.875 \times 10^{-26} \)
So, \( N = \frac{3.2 \times 10^{-8}}{19.875 \times 10^{-26}} \)
\( N \approx 0.161 \times 10^{18} \)
\( N \approx 1.61 \times 10^{17} \text{ s}^{-1} \)
This means the laser emits approximately \( 1.61 \times 10^{17} \) photons every second.
In simple words: To find how many light particles (photons) a laser sends out each second, we use its power and the color of its light. We divide the laser's total power by the energy of a single light particle.

🎯 Exam Tip: Ensure you use consistent units (SI units are preferred) for all quantities, especially for Planck's constant (Js) and wavelength (m), to avoid errors in calculation.

Question 2. Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment.
Answer:
Given:
Stopping Potential \( V_0 = 81 \text{ V} \)
We need to find the maximum kinetic energy \( KE_{max} \) and maximum velocity \( v_{max} \) of the photoelectrons.
The maximum kinetic energy of the emitted photoelectrons is related to the stopping potential by the equation:
\( KE_{max} = eV_0 \)
Where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \text{ C} \)).
So, \( KE_{max} = 1.6 \times 10^{-19} \text{ C} \times 81 \text{ V} \)
\( KE_{max} = 129.6 \times 10^{-19} \text{ J} \)
We can also express this in a more standard form:
\( KE_{max} = 1.296 \times 10^{-17} \text{ J} \)
Now, to find the maximum velocity, we use the kinetic energy formula:
\( KE_{max} = \frac{1}{2} mv_{max}^2 \)
Where \( m \) is the mass of an electron (\( 9.1 \times 10^{-31} \text{ kg} \)).
Rearranging for \( v_{max} \):
\( v_{max} = \sqrt{\frac{2 \times KE_{max}}{m}} \)
\( v_{max} = \sqrt{\frac{2 \times 129.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \)
\( v_{max} = \sqrt{\frac{259.2 \times 10^{-19}}{9.1 \times 10^{-31}}} \)
\( v_{max} = \sqrt{28.48 \times 10^{12}} \)
\( v_{max} \approx 5.33 \times 10^6 \text{ m/s} \)
This shows that even a relatively small stopping potential can lead to electrons moving at very high speeds, demonstrating the significant energy they acquire.
In simple words: If we know the voltage needed to stop the fastest electrons in a photoelectric experiment, we can find out how much energy they had and how fast they were moving.

🎯 Exam Tip: Remember that the maximum kinetic energy of photoelectrons is directly proportional to the stopping potential, and it equals \( eV_0 \).

Question 3. Calculate the energies of the photons associated with the following radiation:
(i) violet light of 413 nm,
(ii) X-rays of 0.1 nm,
(iii) radio waves of 10 m
Answer:
The energy of a photon \( E \) can be calculated using the formula \( E = hv \) or \( E = \frac{hc}{\lambda} \). We will calculate the energy in Joules and then convert it to electronvolts (eV) for better comparison.
(Constants: \( h = 6.625 \times 10^{-34} \text{ Js} \), \( c = 3 \times 10^8 \text{ m/s} \), \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \)).
We can also use the convenient formula for energy in eV: \( E (\text{eV}) = \frac{1240}{\lambda (\text{nm})} \) or \( E (\text{eV}) = \frac{12400}{\lambda (\text{Å})} \). Let's use the first method for consistency.
We can combine \( hc \) into a single value: \( hc = 6.625 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s} = 19.875 \times 10^{-26} \text{ Jm} \).
(i) **Violet light of 413 nm:**
\( \lambda = 413 \text{ nm} = 413 \times 10^{-9} \text{ m} \)
\( E = \frac{hc}{\lambda} = \frac{19.875 \times 10^{-26} \text{ Jm}}{413 \times 10^{-9} \text{ m}} \)
\( E = \frac{19.875}{413} \times 10^{-17} \text{ J} \)
\( E \approx 0.04812 \times 10^{-17} \text{ J} = 4.812 \times 10^{-19} \text{ J} \)
Converting to eV: \( E (\text{eV}) = \frac{4.812 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 3.0075 \text{ eV} \)
So, \( E \approx 3 \text{ eV} \)
(ii) **X-rays of 0.1 nm:**
\( \lambda = 0.1 \text{ nm} = 0.1 \times 10^{-9} \text{ m} = 1 \times 10^{-10} \text{ m} \)
\( E = \frac{hc}{\lambda} = \frac{19.875 \times 10^{-26} \text{ Jm}}{1 \times 10^{-10} \text{ m}} \)
\( E = 19.875 \times 10^{-16} \text{ J} \)
Converting to eV: \( E (\text{eV}) = \frac{19.875 \times 10^{-16} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \)
\( E \approx 12.42 \times 10^3 \text{ eV} = 12420 \text{ eV} \)
This high energy is characteristic of X-rays.
(iii) **Radio waves of 10 m:**
\( \lambda = 10 \text{ m} \)
\( E = \frac{hc}{\lambda} = \frac{19.875 \times 10^{-26} \text{ Jm}}{10 \text{ m}} \)
\( E = 1.9875 \times 10^{-27} \text{ J} \)
Converting to eV: \( E (\text{eV}) = \frac{1.9875 \times 10^{-27} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \)
\( E \approx 1.242 \times 10^{-8} \text{ eV} \)
As expected, radio waves have very low photon energy compared to visible light and X-rays.
In simple words: We find the energy of light particles (photons) using their wavelength. Shorter wavelengths, like X-rays, have much higher energy, while longer wavelengths, like radio waves, have very little energy.

🎯 Exam Tip: Remember that energy and wavelength are inversely proportional. Always convert wavelength to meters and use Planck's constant in Js for energy in Joules, or use the direct eV conversion formula for quick checks.

Question 4. A 150 W lamp emits light of mean wavelength of 5500 Å. If the efficiency is 12%, find out the number of photons emitted by the lamp in one second.
Answer:
Given:
Power of lamp \( P = 150 \text{ W} \)
Mean wavelength \( \lambda = 5500 \text{ Å} = 5500 \times 10^{-10} \text{ m} \)
Efficiency \( Eff = 12\% \)
First, we calculate the effective power used for light emission, considering the efficiency.
Effective power \( P_{eff} = P \times \text{efficiency} = 150 \text{ W} \times \frac{12}{100} = 18 \text{ W} \).
Now, we calculate the energy of a single photon emitted:
\( E = \frac{hc}{\lambda} \)
Using \( h = 6.626 \times 10^{-34} \text{ Js} \) and \( c = 3 \times 10^8 \text{ m/s} \):
\( E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5500 \times 10^{-10}} \)
\( E = \frac{19.878 \times 10^{-26}}{5500 \times 10^{-10}} \)
\( E = \frac{19.878}{5500} \times 10^{-16} \text{ J} \)
\( E \approx 0.003614 \times 10^{-16} \text{ J} = 3.614 \times 10^{-19} \text{ J} \)
The number of photons emitted per second \( N \) is the effective power divided by the energy of one photon:
\( N = \frac{P_{eff}}{E} = \frac{18 \text{ W}}{3.614 \times 10^{-19} \text{ J}} \)
\( N \approx 4.98 \times 10^{19} \text{ photons/second} \)
This large number highlights how many individual energy packets (photons) are involved even in a seemingly simple light source.
In simple words: To find how many light particles a lamp sends out each second, we first figure out how much power it actually uses to make light, considering its efficiency. Then, we divide that useful power by the energy of a single light particle.

🎯 Exam Tip: Always account for the efficiency of the source when calculating the actual power converted into light, as only this power contributes to photon emission.

Question 5. How many photons of frequency 10\(^{14}\) Hz will make up 19.86 J of energy?
Answer:
Given:
Frequency \( \nu = 10^{14} \text{ Hz} \)
Total energy \( E_{total} = 19.86 \text{ J} \)
We need to find the number of photons \( N \).
First, we calculate the energy of a single photon using Planck's formula \( E_{photon} = h\nu \).
Using Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \):
\( E_{photon} = (6.626 \times 10^{-34} \text{ Js}) \times (10^{14} \text{ Hz}) \)
\( E_{photon} = 6.626 \times 10^{-20} \text{ J} \)
The total energy \( E_{total} \) is the number of photons \( N \) multiplied by the energy of one photon \( E_{photon} \).
So, \( E_{total} = N \times E_{photon} \)
Rearranging to find \( N \):
\( N = \frac{E_{total}}{E_{photon}} \)
\( N = \frac{19.86 \text{ J}}{6.626 \times 10^{-20} \text{ J}} \)
\( N = \frac{19.86}{6.626} \times 10^{20} \)
\( N \approx 2.997 \times 10^{20} \)
Rounding to a whole number, the number of photons \( N \approx 3 \times 10^{20} \). This calculation demonstrates how even small amounts of macroscopic energy contain an enormous number of photons.
In simple words: To find how many light particles (photons) are in a certain amount of energy, we divide the total energy by the energy of just one light particle. The energy of one light particle depends on its frequency.

🎯 Exam Tip: Always remember that the energy of a single photon is directly proportional to its frequency (E=hν), which is fundamental to quantum physics.

Question 6. What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
Answer:
Given:
Wavelength of photon \( \lambda = 4000 \text{ Å} = 4000 \times 10^{-10} \text{ m} \)
We need to find the velocity \( v_e \) of the electron such that its momentum equals that of the photon.
First, calculate the momentum of the photon. The momentum of a photon is given by:
\( p_{photon} = \frac{h}{\lambda} \)
Using Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \):
\( p_{photon} = \frac{6.626 \times 10^{-34} \text{ Js}}{4000 \times 10^{-10} \text{ m}} \)
\( p_{photon} = \frac{6.626}{4000} \times 10^{-24} \text{ kg m/s} \)
\( p_{photon} = 1.6565 \times 10^{-27} \text{ kg m/s} \)
Now, we set the momentum of the electron equal to the momentum of the photon:
\( p_{electron} = p_{photon} \)
\( m_e v_e = 1.6565 \times 10^{-27} \text{ kg m/s} \)
Where \( m_e \) is the mass of an electron (\( 9.11 \times 10^{-31} \text{ kg} \)).
Solving for \( v_e \):
\( v_e = \frac{1.6565 \times 10^{-27} \text{ kg m/s}}{9.11 \times 10^{-31} \text{ kg}} \)
\( v_e = \frac{1.6565}{9.11} \times 10^4 \text{ m/s} \)
\( v_e \approx 0.1818 \times 10^4 \text{ m/s} \)
\( v_e \approx 1818 \text{ m/s} \)
This velocity is relatively slow compared to the speed of light, showing how different the momentum relationship is for photons versus particles with mass.
In simple words: We want to find how fast an electron needs to move to have the same "pushing power" (momentum) as a photon of a certain color. We first calculate the photon's momentum, then use the electron's mass to find the speed needed to match that momentum.

🎯 Exam Tip: Remember to distinguish between the momentum of a photon (\( p = h/\lambda \)) and the momentum of a particle with mass (\( p = mv \)).

Question 7. When a light of frequency 9 \( \times \) 10\(^{14}\) Hz is incident on a metal surface, photoelectrons are emitted with a maximum speed of 8 \( \times \) 10\(^5\) ms\(^{-1}\). Determine the threshold frequency of the surface.
Answer:
Given:
Incident light frequency \( \nu = 9 \times 10^{14} \text{ Hz} \)
Maximum speed of photoelectrons \( v_{max} = 8 \times 10^5 \text{ m/s} \)
We need to determine the threshold frequency \( \nu_0 \).
According to Einstein's photoelectric equation:
\( h\nu = h\nu_0 + \frac{1}{2} m_e v_{max}^2 \)
Where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ Js} \)) and \( m_e \) is the mass of an electron (\( 9.11 \times 10^{-31} \text{ kg} \)).
First, calculate the maximum kinetic energy (\( KE_{max} \)) of the photoelectrons:
\( KE_{max} = \frac{1}{2} m_e v_{max}^2 \)
\( KE_{max} = \frac{1}{2} \times (9.11 \times 10^{-31} \text{ kg}) \times (8 \times 10^5 \text{ m/s})^2 \)
\( KE_{max} = \frac{1}{2} \times 9.11 \times 10^{-31} \times 64 \times 10^{10} \)
\( KE_{max} = 9.11 \times 32 \times 10^{-21} \)
\( KE_{max} = 291.52 \times 10^{-21} \text{ J} = 2.9152 \times 10^{-19} \text{ J} \)
Now, rearrange Einstein's equation to solve for \( h\nu_0 \):
\( h\nu_0 = h\nu - KE_{max} \)
\( h\nu_0 = (6.626 \times 10^{-34} \text{ Js}) \times (9 \times 10^{14} \text{ Hz}) - 2.9152 \times 10^{-19} \text{ J} \)
\( h\nu_0 = 59.634 \times 10^{-20} \text{ J} - 2.9152 \times 10^{-19} \text{ J} \)
\( h\nu_0 = 5.9634 \times 10^{-19} \text{ J} - 2.9152 \times 10^{-19} \text{ J} \)
\( h\nu_0 = (5.9634 - 2.9152) \times 10^{-19} \text{ J} \)
\( h\nu_0 = 3.0482 \times 10^{-19} \text{ J} \)
Finally, solve for \( \nu_0 \):
\( \nu_0 = \frac{3.0482 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ Js}} \)
\( \nu_0 = \frac{3.0482}{6.626} \times 10^{15} \text{ Hz} \)
\( \nu_0 \approx 0.460 \times 10^{15} \text{ Hz} = 4.60 \times 10^{14} \text{ Hz} \)
The threshold frequency is a critical property of the metal, indicating the minimum light frequency required for electrons to be ejected.
In simple words: We used Einstein's rule for the photoelectric effect. We calculated the energy of the incoming light and the kinetic energy of the fastest emitted electrons. By subtracting the electron's kinetic energy from the light's energy, we found the energy needed to just free an electron, which then helped us find the threshold frequency.

🎯 Exam Tip: Always make sure to calculate the kinetic energy of the emitted electrons first, then use Einstein's photoelectric equation to find the work function or threshold frequency.

Question 8. When a 6000 Å light falls on the cathode of a photocell and produced photoemission. if a stopping potential of 0.8 V is required to stop the emission of an electron, then determine the
(i) frequency of the light
(ii) the energy of the incident photon
(iii) the work function of the cathode material
(iv) threshold frequency and
(v) the net energy of the electron after it leaves the surface.
Answer:
Given:
Wavelength of incident light \( \lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m} \)
Stopping potential \( V_0 = 0.8 \text{ V} \)
(i) **Frequency of the light \( \nu \):**
We use the formula \( \nu = \frac{c}{\lambda} \).
\( \nu = \frac{3 \times 10^8 \text{ m/s}}{6 \times 10^{-7} \text{ m}} \)
\( \nu = 0.5 \times 10^{15} \text{ Hz} = 5 \times 10^{14} \text{ Hz} \)
(ii) **Energy of the incident photon \( E \):**
\( E = h\nu \)
\( E = (6.626 \times 10^{-34} \text{ Js}) \times (5 \times 10^{14} \text{ Hz}) \)
\( E = 33.13 \times 10^{-20} \text{ J} = 3.313 \times 10^{-19} \text{ J} \)
Alternatively, we can express this in electronvolts (eV):
\( E (\text{eV}) = \frac{3.313 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 2.07 \text{ eV} \)
(iii) **Work function of the cathode material \( W \):**
The maximum kinetic energy of the emitted photoelectrons is \( KE_{max} = eV_0 \).
\( KE_{max} = (1.6 \times 10^{-19} \text{ C}) \times (0.8 \text{ V}) = 1.28 \times 10^{-19} \text{ J} \)
In eV, \( KE_{max} = 0.8 \text{ eV} \).
From Einstein's photoelectric equation, \( E = W + KE_{max} \), so \( W = E - KE_{max} \).
\( W = (3.313 \times 10^{-19} \text{ J}) - (1.28 \times 10^{-19} \text{ J}) \)
\( W = 2.033 \times 10^{-19} \text{ J} \)
In eV, \( W = 2.07 \text{ eV} - 0.8 \text{ eV} = 1.27 \text{ eV} \). The work function is the minimum energy needed to free an electron.
(iv) **Threshold frequency \( \nu_0 \):**
We use \( W = h\nu_0 \), so \( \nu_0 = \frac{W}{h} \).
\( \nu_0 = \frac{2.033 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ Js}} \)
\( \nu_0 \approx 0.3068 \times 10^{15} \text{ Hz} = 3.068 \times 10^{14} \text{ Hz} \)
(v) **Net energy of the electron after it leaves the surface:**
This refers to the maximum kinetic energy of the emitted electron.
Net energy \( = KE_{max} = E - W \)
Net energy \( = 2.07 \text{ eV} - 1.27 \text{ eV} = 0.8 \text{ eV} \)
This confirms that the kinetic energy is directly related to the stopping potential.
In simple words: We used the light's color and the stopping voltage to find several important values for the photocell. We found the light's frequency and energy, how much energy is needed to just release an electron (work function), the minimum frequency required to do this (threshold frequency), and the fastest an electron can move.

🎯 Exam Tip: Systematically break down multi-part problems into individual calculations, ensuring you use the correct formula and units for each step (Joules or electronvolts as appropriate).

Question 9. A 3310 Å photon liberates an electron from a material with energy 3 \( \times \) 10\(^{-19}\) J while another 5000 Å photon ejects an electron with energy 0.972 \( \times \) 10\(^{-19}\) J from the same material. Determine the value of Planck's constant and the threshold wavelength of the material.
Answer:
Given:
For the first photon:
Wavelength \( \lambda_1 = 3310 \text{ Å} = 3310 \times 10^{-10} \text{ m} \)
Kinetic energy of emitted electron \( KE_1 = 3 \times 10^{-19} \text{ J} \)
For the second photon:
Wavelength \( \lambda_2 = 5000 \text{ Å} = 5000 \times 10^{-10} \text{ m} \)
Kinetic energy of emitted electron \( KE_2 = 0.972 \times 10^{-19} \text{ J} \)

First, calculate the frequencies of the incident photons:
Frequency \( \nu_1 = \frac{c}{\lambda_1} = \frac{3 \times 10^8 \text{ m/s}}{3310 \times 10^{-10} \text{ m}} \approx 9.06 \times 10^{14} \text{ Hz} \)
Frequency \( \nu_2 = \frac{c}{\lambda_2} = \frac{3 \times 10^8 \text{ m/s}}{5000 \times 10^{-10} \text{ m}} = 6 \times 10^{14} \text{ Hz} \)
According to Einstein's photoelectric equation, for a given material, the work function \( W \) is constant:
\( h\nu_1 = W + KE_1 \) ..............(1)
\( h\nu_2 = W + KE_2 \) ..............(2)
Subtract equation (2) from equation (1) to eliminate \( W \):
\( h\nu_1 - h\nu_2 = (W + KE_1) - (W + KE_2) \)
\( h(\nu_1 - \nu_2) = KE_1 - KE_2 \)
Now, substitute the known values to find Planck's constant \( h \):
\( h(9.06 \times 10^{14} - 6 \times 10^{14}) = (3 \times 10^{-19} - 0.972 \times 10^{-19}) \)
\( h(3.06 \times 10^{14}) = 2.028 \times 10^{-19} \)
\( h = \frac{2.028 \times 10^{-19}}{3.06 \times 10^{14}} \)
\( h \approx 0.6626 \times 10^{-33} \text{ Js} \)
So, Planck's constant \( h \approx 6.626 \times 10^{-34} \text{ Js} \). This calculation is a practical way to determine fundamental physical constants.
Next, we find the work function \( W \) using equation (1) and the calculated \( h \):
\( W = h\nu_1 - KE_1 \)
\( W = (6.626 \times 10^{-34} \text{ Js}) \times (9.06 \times 10^{14} \text{ Hz}) - (3 \times 10^{-19} \text{ J}) \)
\( W = 5.991 \times 10^{-19} \text{ J} - 3 \times 10^{-19} \text{ J} \)
\( W = 2.991 \times 10^{-19} \text{ J} \)
Finally, calculate the threshold wavelength \( \lambda_0 \) using \( W = \frac{hc}{\lambda_0} \):
\( \lambda_0 = \frac{hc}{W} \)
\( \lambda_0 = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{2.991 \times 10^{-19} \text{ J}} \)
\( \lambda_0 = \frac{19.878 \times 10^{-26}}{2.991 \times 10^{-19}} \)
\( \lambda_0 \approx 6.646 \times 10^{-7} \text{ m} \)
\( \lambda_0 \approx 6646 \text{ Å} \)
In simple words: By comparing the energy of different light colors and the energy of the electrons they release from a metal, we can figure out two key numbers: Planck's constant, which is a basic constant of nature, and the longest wavelength of light that can still release an electron from that metal.

🎯 Exam Tip: Remember that the work function (W) of a given material is constant. This allows you to set up simultaneous equations from different incident photon energies to solve for unknowns like Planck's constant or the work function itself.

Question 10. At the given point of time, the earth receives energy from the sun at 4 cal cm\(^{-2}\) min\(^{-1}\). Determine the number of photons received on the surface of the Earth per cm\(^{-2}\) per minute. (Given: Mean wavelength of sunlight = 5500 Å)
Answer:
Given:
Energy received per unit area per unit time \( E = 4 \text{ cal cm}^{-2} \text{ min}^{-1} \)
Mean wavelength of sunlight \( \lambda = 5500 \text{ Å} = 5500 \times 10^{-10} \text{ m} \)
We need to find the number of photons received per cm\(^{-2}\) per minute.
First, convert the energy from calories to Joules. We know that \( 1 \text{ cal} = 4.2 \text{ J} \).
So, \( E = 4 \text{ cal} \times 4.2 \text{ J/cal} = 16.8 \text{ J cm}^{-2} \text{ min}^{-1} \).
Next, calculate the energy of a single photon of mean sunlight wavelength:
\( E_{photon} = \frac{hc}{\lambda} \)
Using Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \) and speed of light \( c = 3 \times 10^8 \text{ m/s} \):
\( E_{photon} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5500 \times 10^{-10}} \)
\( E_{photon} = \frac{19.878 \times 10^{-26}}{5500 \times 10^{-10}} \)
\( E_{photon} \approx 0.003614 \times 10^{-16} \text{ J} = 3.614 \times 10^{-19} \text{ J} \)
The number of photons \( N \) received per cm\(^{-2}\) per minute is the total energy received per cm\(^{-2}\) per minute divided by the energy of a single photon:
\( N = \frac{E}{E_{photon}} \)
\( N = \frac{16.8 \text{ J}}{3.614 \times 10^{-19} \text{ J/photon}} \)
\( N \approx 4.648 \times 10^{19} \text{ photons cm}^{-2} \text{ min}^{-1} \)
This incredibly large number of photons reaching Earth illustrates the vast energy output of the sun and the discrete nature of light.
In simple words: We find out how many light particles from the sun hit a small area of Earth in one minute. We do this by converting the total energy received into Joules, then dividing it by the energy of a single light particle from the sun.

🎯 Exam Tip: Pay close attention to unit conversions (calories to Joules, Ångstroms to meters) as these are common sources of error in such problems.

Question 11. UV light of wavelength 1800 Å is incident on a lithium surface whose threshold wavelength 4965 Å. Determine the maximum energy of the electron emitted.
Answer:
Given:
Wavelength of incident UV light \( \lambda = 1800 \text{ Å} = 1800 \times 10^{-10} \text{ m} \)
Threshold wavelength of lithium \( \lambda_0 = 4965 \text{ Å} = 4965 \times 10^{-10} \text{ m} \)
We need to determine the maximum kinetic energy \( KE_{max} \) of the emitted electron.
According to Einstein's photoelectric equation:
\( KE_{max} = E - W \)
Where \( E \) is the energy of the incident photon (\( E = \frac{hc}{\lambda} \)) and \( W \) is the work function of the material (\( W = \frac{hc}{\lambda_0} \)).
So, \( KE_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \)
Using Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \) and speed of light \( c = 3 \times 10^8 \text{ m/s} \):
\( hc = 19.878 \times 10^{-26} \text{ Jm} \)
\( KE_{max} = 19.878 \times 10^{-26} \left( \frac{1}{1800 \times 10^{-10}} - \frac{1}{4965 \times 10^{-10}} \right) \)
\( KE_{max} = 19.878 \times 10^{-26} \times \frac{1}{10^{-10}} \left( \frac{1}{1800} - \frac{1}{4965} \right) \)
\( KE_{max} = 19.878 \times 10^{-16} \left( \frac{4965 - 1800}{1800 \times 4965} \right) \)
\( KE_{max} = 19.878 \times 10^{-16} \left( \frac{3165}{8937000} \right) \)
\( KE_{max} \approx 19.878 \times 10^{-16} \times 3.541 \times 10^{-4} \)
\( KE_{max} \approx 70.36 \times 10^{-20} \text{ J} = 7.036 \times 10^{-19} \text{ J} \)
To express this in electronvolts (eV), divide by \( 1.6 \times 10^{-19} \text{ J/eV} \):
\( KE_{max} (\text{eV}) = \frac{7.036 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 4.3975 \text{ eV} \)
So, \( KE_{max} \approx 4.4 \text{ eV} \). This calculation is vital in understanding the energy balance in the photoelectric effect.
In simple words: To find the maximum energy of an electron released from a metal, we subtract the energy needed to free the electron (work function) from the energy of the incoming light. Both these energies depend on their respective wavelengths.

🎯 Exam Tip: Be careful with the units for wavelength (Å or m) and ensure consistency throughout the calculation. The formula \( KE_{max} = hc(\frac{1}{\lambda} - \frac{1}{\lambda_0}) \) is very useful here.

Question 12. Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to 81.9 \( \times \) 10\(^{-15}\) J. (Given: mass of proton is 1836 times that of electron).
Answer:
Given:
Kinetic energy of proton \( KE = 81.9 \times 10^{-15} \text{ J} \)
Mass of proton \( m_p = 1836 \times m_e \) (where \( m_e = 9.11 \times 10^{-31} \text{ kg} \))
\( m_p = 1836 \times 9.11 \times 10^{-31} \text{ kg} \approx 1.673 \times 10^{-27} \text{ kg} \)
We need to calculate the de Broglie wavelength \( \lambda \).
The de Broglie wavelength is given by the formula:
\( \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}} \)
Using Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \):
\( \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times (1.673 \times 10^{-27} \text{ kg}) \times (81.9 \times 10^{-15} \text{ J})}} \)
First, calculate the term inside the square root:
\( 2 \times 1.673 \times 10^{-27} \times 81.9 \times 10^{-15} \)
\( = 274.0534 \times 10^{-42} \)
So, \( \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{274.0534 \times 10^{-42}}} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{16.554 \times 10^{-21}} \)
\( \lambda \approx 0.4002 \times 10^{-13} \text{ m} \)
\( \lambda \approx 4 \times 10^{-14} \text{ m} \)
This wavelength is extremely small, as expected for a particle with significant mass and kinetic energy, highlighting the wave-particle duality at the subatomic level.
In simple words: To find the wave-like length of a proton, we use its energy and its mass. We use a special formula that includes Planck's constant to relate the proton's energy and mass to its de Broglie wavelength.

🎯 Exam Tip: Make sure to use the correct mass for the particle (proton in this case) and convert all units to SI before calculation. Remember the direct relationship between wavelength, mass, and kinetic energy: \( \lambda \propto \frac{1}{\sqrt{mKE}} \).

Question 13. A deuteron and an alpha particle are accelerated with the same potential. Which one of the two has
(i) greater value of de Broglie wavelength associated with it and
(ii) less kinetic energy? Explain.
Answer:
Given: A deuteron and an alpha particle are accelerated by the same potential \( V \).
Let's analyze their properties:
**Deuteron:**
Charge \( q_d = +e \)
Mass \( m_d \approx 2m_p \) (where \( m_p \) is the mass of a proton)
**Alpha Particle:**
Charge \( q_\alpha = +2e \)
Mass \( m_\alpha \approx 4m_p \)

(i) **Greater de Broglie wavelength:**
The de Broglie wavelength \( \lambda \) for a charged particle accelerated through a potential \( V \) is given by:
\( \lambda = \frac{h}{\sqrt{2mqV}} \)
Since \( h \) and \( V \) are constant, \( \lambda \propto \frac{1}{\sqrt{mq}} \).
For a deuteron: \( \lambda_d \propto \frac{1}{\sqrt{m_d q_d}} = \frac{1}{\sqrt{(2m_p)(e)}} = \frac{1}{\sqrt{2m_p e}} \)
For an alpha particle: \( \lambda_\alpha \propto \frac{1}{\sqrt{m_\alpha q_\alpha}} = \frac{1}{\sqrt{(4m_p)(2e)}} = \frac{1}{\sqrt{8m_p e}} \)
Comparing the denominators, \( \sqrt{2m_p e} < \sqrt{8m_p e} \). Therefore, \( \lambda_d > \lambda_\alpha \).
The deuteron has a greater de Broglie wavelength.
This means the wave-like properties are more pronounced for the deuteron under the same acceleration conditions.

(ii) **Less kinetic energy:**
The kinetic energy \( KE \) gained by a charged particle accelerated through a potential \( V \) is given by:
\( KE = qV \)
Since the potential \( V \) is the same for both:
For a deuteron: \( KE_d = q_d V = eV \)
For an alpha particle: \( KE_\alpha = q_\alpha V = 2eV \)
Comparing their kinetic energies, \( KE_d = eV \) and \( KE_\alpha = 2eV \).
Therefore, \( KE_d < KE_\alpha \).
The deuteron has less kinetic energy. The alpha particle has twice the charge, so it gains twice as much kinetic energy when accelerated by the same potential difference. This illustrates how both mass and charge influence a particle's behavior in an electric field.
In simple words: When a deuteron and an alpha particle are sped up by the same voltage, the deuteron will have a longer de Broglie wavelength because it has less charge per unit mass. The deuteron will also have less kinetic energy because it has less charge than the alpha particle, even though the voltage is the same.

🎯 Exam Tip: Remember that kinetic energy gained is \( qV \) and de Broglie wavelength is inversely proportional to \( \sqrt{mqV} \). Analyze the mass and charge of each particle carefully when comparing them.

Question 14. An electron is accelerated through a potential difference of 81 V. What is the de Broglie wavelength associated with it? To which part of electromagnetic spectrum does this wavelength correspond?
Answer:
Given:
Potential difference \( V = 81 \text{ V} \)
We need to find the de Broglie wavelength \( \lambda \) associated with the electron.
The de Broglie wavelength \( \lambda \) for an electron accelerated through a potential \( V \) is given by the formula:
\( \lambda = \frac{12.27}{\sqrt{V}} \text{ Å} \)
Substitute the given potential difference:
\( \lambda = \frac{12.27}{\sqrt{81}} \text{ Å} \)
\( \lambda = \frac{12.27}{9} \text{ Å} \)
\( \lambda \approx 1.363 \text{ Å} \)
So, the de Broglie wavelength associated with the electron is approximately \( 1.36 \text{ Å} \).
This wavelength can also be expressed in meters: \( 1.36 \times 10^{-10} \text{ m} \).
This wavelength falls in the range of **X-rays** in the electromagnetic spectrum. X-rays typically have wavelengths from \( 0.01 \text{ Å} \) to \( 100 \text{ Å} \) (\( 10^{-12} \text{ m} \) to \( 10^{-8} \text{ m} \)). This illustrates how electrons can exhibit wave-like behavior comparable to high-energy electromagnetic radiation.
In simple words: When an electron speeds up due to a voltage, it also acts like a wave. We calculate this wave's length using a specific formula. This wave-like length is so short that it falls into the category of X-rays on the light spectrum.

🎯 Exam Tip: Memorize the simplified formula for electron de Broglie wavelength in Angstroms: \( \lambda = \frac{12.27}{\sqrt{V}} \text{ Å} \), and know the typical wavelength ranges for different parts of the electromagnetic spectrum.

Part II:

I. Match the Following:

Question 1.

🎯 Exam Tip: Remember specific applications for each emission type to accurately match them.

Question 2.

🎯 Exam Tip: Knowing the historical timeline of key physics discoveries helps in matching experiments and theories with their respective years.

Question 3.

🎯 Exam Tip: Understand the primary function and key components of each type of photocell to match them correctly.

Question 4.

🎯 Exam Tip: Link each physicist to their primary contribution or discovery in the field of dual nature of radiation and matter.

II. Fill in the Blanks:

Question 1. The liberation of electrons from any surface of a substance is called __________.
Answer: electron emission
In simple words: When electrons come out from a material's surface, it's called electron emission.

🎯 Exam Tip: Define key terms clearly and precisely in physics. This is a fundamental definition.

Question 2. 1 eV is equal to __________ Joule.
Answer: \( 1.602 \times 10^{-19} \)
In simple words: One electron volt (eV) is a unit of energy, and its value in Joules is a constant.

🎯 Exam Tip: Memorize the conversion factor between electron volts and Joules, as it is often used in problems.

Question 3. The stopping potential is independent of __________ of the incident light.
Answer: intensity
In simple words: The voltage needed to stop electrons (stopping potential) does not change even if the light gets brighter or dimmer (intensity).

🎯 Exam Tip: Understand that stopping potential depends on frequency, not intensity. Intensity affects the number of electrons, not their maximum kinetic energy.

Question 4. The quality of X-rays is measured in terms of their __________.
Answer: penetrating power
In simple words: How good X-rays are at going through things (their quality) is measured by how deep they can go (penetrating power).

🎯 Exam Tip: Relate the "quality" of X-rays to their ability to penetrate materials, which is determined by their energy or wavelength.

Question 5. The graph between maximum kinetic energy of the photoelectron and frequency of the incident light is __________.
Answer: straight line
In simple words: If you plot a graph showing how the fastest electron's energy changes with light frequency, it makes a straight line.

🎯 Exam Tip: Recall Einstein's photoelectric equation which shows a linear relationship between maximum kinetic energy and frequency.

Question 6. Einstein's photoelectric equation was experimentally confirmed by __________.
Answer: Millikan
In simple words: Robert Millikan did experiments that proved Einstein's ideas about the photoelectric effect were correct.

🎯 Exam Tip: Remember Millikan's oil-drop experiment and his confirmation of Einstein's photoelectric equation as key experimental validations.

III. Choose the Odd Man Out:

Question 1.
(a) Field emission display
(b) Photodiodes
(c) Photomultiplier tubes
(d) Electron microscope
Answer: (d) Electron microscope
In simple words: An electron microscope is an imaging device, while the others are related to electron emission or detection.

🎯 Exam Tip: Categorize the given items by their primary function or underlying principle to identify the one that doesn't fit.

Question 2.
(a) Work function
(b) Discharge tube
(c) Threshold frequency
(d) Stopping potential
Answer: (b) Discharge tube
In simple words: Work function, threshold frequency, and stopping potential are all terms related to the photoelectric effect, while a discharge tube is a device used in different contexts.

🎯 Exam Tip: Group terms by the physics phenomenon they describe. The discharge tube operates on gas ionization, distinct from photoelectric effect concepts.

Question 3.
(a) mass
(b) frequency
(c) Intensity
(d) wavelength
Answer: (a) mass
In simple words: Frequency, intensity, and wavelength are properties of light or waves. Mass is a property of matter.

🎯 Exam Tip: Differentiate between properties related to waves and particles. Mass is a particle property, while the others describe waves.

Question 4.
(a) neutrons
(b) tungsten
(c) particles
(d) x-rays
Answer: (a) tungsten
In simple words: Tungsten is a material (metal), while neutrons, particles, and X-rays are types of radiation or matter at a subatomic level.

🎯 Exam Tip: Identify the common category for most items and then find the one that falls outside that category.

Question 5.
(a) Photon
(b) electron
(c) proton
(d) neutron
Answer: (a) Photon
In simple words: Electrons, protons, and neutrons are all particles with mass that make up atoms. A photon is a particle of light (energy) with no mass.

🎯 Exam Tip: Recall the fundamental particles and distinguish between matter particles (leptons and hadrons) and force carrier particles (like photons).

IV. Choose the Incorrect Pair:

Question 1.

🎯 Exam Tip: Understand the properties of continuous and characteristic X-rays, including factors affecting their intensity and wavelength.

Question 2.

🎯 Exam Tip: Review the laws of photoelectric effect. While emission starts instantly, its rate (photocurrent) and stopping potential are influenced by electrode potential.

Question 3.

🎯 Exam Tip: Accurately define key terms in the photoelectric effect. Stopping potential is a retarding potential, specifically the minimum value that halts current.

V. Choose the Correct Pair:

Question 1.

🎯 Exam Tip: Familiarize yourself with standard values and formulas for de-Broglie wavelength and fundamental constants. Remember, the charge of an electron is \( 1.6 \times 10^{-19} \) C, not positive \( 1.6 \times 10^{19} \) C.

Question 2.

🎯 Exam Tip: Carefully check each formula against known physics principles and units to ensure correctness, especially for relationships involving fundamental constants and variables.

VI. Assertion and Reason:

Question 1. Assertion: The resolving power of the electron microscope is high. Reason: The wavelength of electrons is very much lesser than the visible light.
(i) Assertion is true but reason is false
(ii) The assertion and reason both are false
(iii) Both assertion and reason are true and the reason is the correct explanation
(iv) Both assertion and reason are true and the reason is not the correct explanation of the assertion.
Answer: (iii) Both assertion and reason are true and the reason is the correct explanation of the assertion.
In simple words: An electron microscope can see very tiny details because electron waves are much shorter than light waves, making the reason a good explanation for the assertion.

🎯 Exam Tip: Remember that resolving power is inversely proportional to wavelength. Shorter wavelengths, like those of electrons, lead to higher resolving power.

Question 2. Assertion: The oscillators emit or absorb energy in small packets or quanta. Reason: Energy is not discrete, the energy posses wave nature.
(i) The assertion and reason both are false
(ii) The assertion is true but reason is false
(iii) The assertion is false but reason is true
(iv) Both the assertion and reason are true
Answer: (ii) The assertion is true but the reason is false
In simple words: Oscillators do emit energy in tiny packets called quanta (the assertion is true), but energy *is* discrete according to quantum theory, not continuous with wave nature in this context (the reason is false).

🎯 Exam Tip: Grasp the core concept of quantum theory: energy is quantized (discrete), not continuous. This is key to understanding Planck's hypothesis.

VII. Choose the Correct Statement:

Question 1.
(a) Particle cannot be localised in space and time
(b) Wave can be localised in space and time
(c) Black body radiation can be explained by wave nature.
(d) The minimum energy needed for an electron to escape from the metal surface is called the work function.
Answer: (d) The minimum energy needed for an electron to escape from the metal surface is called the work function.
In simple words: The correct statement is the definition of work function, which is the smallest amount of energy an electron needs to leave a metal surface.

🎯 Exam Tip: Precisely recall the definitions of fundamental concepts like work function in the context of the photoelectric effect.

Question 2. __________ent of the particle increases de-Broglie wavelength also increases.
Answer: The provided text is incomplete, but if it implies "momentum", then the statement would be incorrect as de-Broglie wavelength is inversely proportional to momentum. However, based on the provided text, it's difficult to give a definitive answer without the complete sentence. Assuming the question intended to ask for a correct statement related to de-Broglie wavelength.
In simple words: The question text is not complete. However, de-Broglie wavelength gets smaller when a particle's momentum gets bigger.

🎯 Exam Tip: The de-Broglie wavelength formula \( \lambda = h/p \) shows an inverse relationship: as momentum (p) increases, wavelength \( \lambda \) decreases. Ensure you understand this inverse proportionality.

VIII. Choose the Incorrect Statement:

Question 1.
(a) Wave mechanical concept of the atom was based on the de-Broglie hypothesis
(b) Maximum kinetic energy of the photoelectrons varies linearly with the frequency of incident radiation.
(c) The stopping potential of a metal surface is independent of the intensity of the incident radiation.
(d) The graph drawn by taking the frequency of the radiation along the x-axis and stopping potential along the y-axis for a photo-sensitive metal is a parabola.
Answer: (d) The graph drawn by taking the frequency of the radiation along the x-axis and stopping potential along the y-axis for a photo-sensitive metal is a parabola.
In simple words: The graph of stopping potential versus frequency is a straight line, not a curve like a parabola. This statement is incorrect.

🎯 Exam Tip: Remember the graphical representations of photoelectric effect laws. The relationship between stopping potential and frequency is linear.

Question 2.
(a) Photoelectric emission is an instantaneous process.
(b) Maximum kinetic energy of photoelectrons is directly proportional to the frequency of the incident radiation.
(c) The energy of photoelectrons is indirectly proportional to the intensity of the incident radiation.
(d) Photoelectron emission is not possible below a minimum frequency of the incident radiation.
Answer: (c) The energy of photoelectrons is indirectly proportional to the intensity of the incident radiation.
In simple words: The maximum kinetic energy of photoelectrons does not depend on the intensity of light; it depends on the frequency. So, this statement is wrong because it wrongly links energy and intensity.

🎯 Exam Tip: Clearly distinguish between the effects of intensity and frequency in the photoelectric effect. Intensity affects the number of electrons, while frequency affects their maximum kinetic energy.

IX. Choose the Best Answer:

Question 1. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 3 eV fall on it is 4 eV. The stopping potential, in volt, is
(a) 2
(b) 4
(c) 6
(d) 10
Answer: (b) 4
In simple words: The stopping potential is equal to the maximum kinetic energy of the emitted electrons, expressed in volts. So if the max kinetic energy is 4 eV, the stopping potential is 4 V.

🎯 Exam Tip: The stopping potential \( V_0 \) is numerically equal to the maximum kinetic energy \( K_{max} \) when \( K_{max} \) is expressed in electron volts (eV), i.e., \( K_{max} = eV_0 \).

Question 2. The threshold frequency is constant with respect to __________
(a) type of the metal
(b) Velocity of electrons
(c) Voltage applied
(d) intensity of the incident light.
Answer: (a) type of the metal
In simple words: The threshold frequency is a property unique to each material. It only depends on what type of metal it is.

🎯 Exam Tip: Threshold frequency is a characteristic property of the material, not dependent on the incident light's velocity, voltage, or intensity.

Question 3. In a photoemissive cell, the anode is made up of __________
(a) copper
(b) gold
(c) platinum
(d) zinc
Answer: (c) platinum
In simple words: In a photoemissive cell, the anode (the part that collects electrons) is often made from platinum because of its specific electrical properties.

🎯 Exam Tip: Specific materials are chosen for anodes and cathodes in photoemissive cells based on their work function and other electrical characteristics.

Question 4. The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately
(a) 540 nm
(b) 400 nm
(c) 310 nm
(d) 220 nm
Answer: (c) 310 nm
In simple words: Using the formula that links work function to threshold wavelength, we can calculate that for a 4.0 eV work function, the longest wavelength of light that can emit electrons is about 310 nm. This helps us find the "cutoff" point for light that can cause emission.

🎯 Exam Tip: Use the formula \( \lambda_0 = hc/\phi_0 \), where \( \lambda_0 \) is the threshold wavelength, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \phi_0 \) is the work function. Remember to convert eV to Joules or use appropriate constants for eV units.

Question 5. Duane-Hunt law is __________
(a) \( \lambda = \frac{12400}{\mathrm{v}} \) m
(b) \( x = \frac{\mathrm{hc}}{\mathrm{V}} \) m
(c) \( \lambda = \frac{\mathrm{hc}}{\sqrt{\mathrm{E}_{\mathrm{k}}}} \) m
(d) none of the options
Answer: (a) \( \lambda = \frac{12400}{\mathrm{v}} \) m
In simple words: The Duane-Hunt law describes the shortest wavelength (cutoff wavelength) of X-rays produced in an X-ray tube. This is determined by the accelerating voltage V, usually given as \( \lambda_{min} = \frac{hc}{eV} \), which can be simplified numerically to \( \frac{12400}{V} \) Å or similar values depending on units.

🎯 Exam Tip: The Duane-Hunt law directly relates the minimum wavelength of X-rays to the accelerating voltage in an X-ray tube, often expressed as \( \lambda_{min} = \frac{12400}{V} \) when \( \lambda \) is in Å and \( V \) in volts.

Question 6. The resolving power of the electron microscope is __________ times greater than the resolving power of the optical microscope
(a) \( 10^2 \)
(b) \( 10^4 \)
(c) \( 10^5 \)
(d) \( 10^3 \)
Answer: (c) \( 10^5 \)
In simple words: An electron microscope can see details about a hundred thousand times smaller than an optical microscope. This is because electron waves are much, much shorter than light waves.

🎯 Exam Tip: The resolving power of a microscope is inversely proportional to the wavelength used. Since electron wavelengths are significantly smaller than visible light wavelengths, electron microscopes have much higher resolving power.

Question 7. 4 eV is the energy of the incident photon and the work function is 2 eV. The stopping potential will be
(a) 2V
(b) 4V
(c) 6V
(d) \( 2\sqrt{2}V \)
Answer: (a) 2V
In simple words: First, find the maximum kinetic energy by subtracting the work function from the photon's energy (4 eV - 2 eV = 2 eV). Then, the stopping potential needed to halt these electrons is numerically equal to this kinetic energy in volts, which is 2 V.

🎯 Exam Tip: Apply Einstein's photoelectric equation: \( K_{max} = E_{photon} - \phi_0 \). The stopping potential \( V_0 \) is related by \( K_{max} = eV_0 \), so if \( K_{max} \) is in eV, then \( V_0 \) is numerically the same value in volts.

Question 8. Which of the following graph represents the variation of the particle momentum and associated de Broglie wavelength.
(a)
\( P \)
\( \lambda \)
(b)
\( P \)
\( \lambda \)
(c)
\( P \)
\( \lambda \)
(d)
\( P \)
\( \lambda \)
Answer: (c)
\( P \)
\( \lambda \)
In simple words: The de Broglie wavelength \( \lambda \) and particle momentum \( P \) are inversely related (\( \lambda = h/P \)). This means as one increases, the other decreases, following a hyperbolic curve. The graph in option (c) correctly shows this inverse relationship.

🎯 Exam Tip: Remember the inverse relationship in de Broglie's hypothesis: \( \lambda \propto 1/P \). This type of relationship is represented by a hyperbola on a graph, not a straight line or exponential curve.

Question 9. The radiation produced from decelerating electron is called
(a) reverse radiation
(b) Photon emission
(c) black body radiation
(d) breaking radiation
Answer: (d) breaking radiation
In simple words: When electrons slow down quickly, they give off a type of radiation called breaking radiation or bremsstrahlung. This happens when fast-moving electrons hit a target and lose energy.

🎯 Exam Tip: Remember that "breaking radiation" (bremsstrahlung) is a key concept when discussing X-ray production and electron deceleration.

Question 10. If the striking electrons knocks of an electron in n = 2 state, the resulting transition give
(a) K-lines
(b) L-lines
(c) M-lines
(d) N-lines
Answer: (b) L-lines
In simple words: When an electron is removed from the second energy level (n=2, which is the L-shell) of an atom and another electron falls into that empty space, the radiation produced is called L-line radiation. It is part of the characteristic X-ray spectrum.

🎯 Exam Tip: Familiarize yourself with electron shell notation (K, L, M, N) and the spectral lines (K-alpha, L-beta, etc.) that correspond to electron transitions between these shells.

Question 11. When a proton is accelerated through 1V, then its kinetic energy will be
(a) 1 eV
(b) 13.6 eV
(c) 1840 eV
(d) 0.54 eV
Answer: (a) 1 eV
In simple words: When a proton, which has a charge equal to an electron, is sped up by a voltage of 1 Volt, it gains a specific amount of energy. This energy is exactly 1 electron volt, regardless of the particle's mass.

🎯 Exam Tip: The kinetic energy gained by any charged particle accelerating through a potential difference V is given by \( KE = qV \). For an elementary charge (like a proton or electron) accelerated through 1 Volt, the energy gained is 1 eV.

Question 12. The work function of a photoelectric material is 3.3 eV. The threshold frequency will be equal to
(a) \( 8 \times 10^{14} \) Hz
(b) \( 8 \times 10^{10} \) Hz
(c) \( 5 \times 10^{20} \) Hz
(d) \( 4 \times 10^{14} \) Hz
Answer: (a) \( 8 \times 10^{14} \) Hz
In simple words: The work function tells us the minimum energy needed to free an electron from a metal surface. We can find the threshold frequency by dividing the work function by Planck's constant, which converts the energy into a frequency value.

🎯 Exam Tip: Remember the relationship \( W = h\nu_0 \), where W is the work function, h is Planck's constant, and \( \nu_0 \) is the threshold frequency. Make sure to use consistent units for energy (Joules or eV) and Planck's constant.

Question 13. The momentum of the electron having wavelength of 4 Å is
(a) \( 1.6 \times 10^{24} \) kg ms⁻¹
(b) \( 1.65 \times 10^{-24} \) kg ms⁻¹
(c) \( 3.3 \times 10^{24} \) kg ms⁻¹
(d) \( 3.3 \times 10^{-24} \) kg ms⁻¹
Answer: (b) \( 1.65 \times 10^{-24} \) kg ms⁻¹
In simple words: To find the electron's momentum from its wavelength, we use de Broglie's formula. This formula connects the wave-like property (wavelength) to the particle-like property (momentum) using Planck's constant.

🎯 Exam Tip: Convert Ångströms (Å) to meters (\( 1 Å = 10^{-10} m \)) before using de Broglie's wavelength formula \( p = \frac{h}{\lambda} \). Pay attention to the negative exponent in the final answer.

Question 15. The number of photo-electrons emitted for the light of a frequency \( \nu \) (higher than the threshold frequency \( \nu_0 \)) is proportional to
(a) Threshold frequency (\( \nu_0 \))
(b) Intensity of light
(c) Frequency of light (\( \nu \))
(d) \( \nu - \nu_0 \)
Answer: (b) Intensity of light
In simple words: In the photoelectric effect, if light has enough energy, more intense light means more photons. Each photon can knock out one electron, so higher intensity leads to more emitted electrons.

🎯 Exam Tip: Remember that the number of photoelectrons depends on the intensity of light (number of photons), while their maximum kinetic energy depends on the frequency of light (energy per photon).

Question 15. The maximum kinetic energy of photoelectrons emitted in a photoelectric effect phenomenon is 3.2 eV. What is the value of stopping potential to stop the photoelectrons?
(a) 1.6 eV
(b) -3.2 eV
(c) -1.6 eV
(d) -6.4 eV
Answer: (b) -3.2 eV
In simple words: The stopping potential is the negative voltage that completely stops the most energetic photoelectrons. Its value is numerically equal to the maximum kinetic energy, but it's negative because it works against the electron's motion.

🎯 Exam Tip: The stopping potential \( V_0 \) is related to the maximum kinetic energy \( KE_{max} \) by the equation \( KE_{max} = eV_0 \). Note that the stopping potential itself is a voltage, and the energy value here is equivalent to the energy needed to stop the electron, hence the negative sign for potential.

Question 16. Radiation of energy 6.2 eV is incident on the metal surface of work function 4.7 eV. Find the kinetic energy of the electrons emitted.
(a) \( 3 \times 10^{-19} \) J
(b) \( 1.6 \times 10^{-19} \) J
(c) \( 2.4 \times 10^{-19} \) J
(d) \( 3.2 \times 10^{-19} \) J
Answer: (c) \( 2.4 \times 10^{-19} \) J
In simple words: We can find the kinetic energy of emitted electrons by subtracting the work function (energy needed to escape) from the incident photon's energy. The remaining energy is what the electron uses to move.

🎯 Exam Tip: Use Einstein's photoelectric equation: \( KE_{max} = h\nu - W \), where \( h\nu \) is the incident photon energy and W is the work function. Convert eV to Joules (\( 1 eV = 1.6 \times 10^{-19} J \)) for the final answer if required in Joules.

Question 17. The de-Broglie wavelength of electrons moving with a speed of 500 km/s is
(a) 1.45 nm
(b) 1.45 Å
(c) 1.45 m
(d) \( 4.6 \times 10^{24} \) Hz
Answer: (a) 1.45 nm
In simple words: We calculate the de-Broglie wavelength by dividing Planck's constant by the electron's momentum (mass times velocity). This shows the wave-like nature of the moving electron.

🎯 Exam Tip: Ensure all units are in SI (meters, kilograms, seconds) before calculation. Remember to convert km/s to m/s and express the final wavelength in nanometers (nm) or Ångströms (Å) as needed for clarity.

Question 18. Electron volt is a unit of
(a) Energy
(b) potential
(c) current
(d) charge
Answer: (a) Energy
In simple words: An electron volt is a small unit of energy. It's the amount of energy one electron gains when it moves through an electric potential difference of one volt.

🎯 Exam Tip: Always associate the term "electron volt" with energy. It's commonly used in atomic and nuclear physics to express very small energy values, and is equivalent to \( 1.602 \times 10^{-19} \) Joules.

Question 19. If the work function of iron is 4.7 eV, calculate the cut-off wavelength for this metal.
(a) 2633 Å
(b) \( 26.4 \times 10^{-7} \) m
(c) \( 13.2 \times 10^{-10} \) m
(d) 1320 Å
Answer: (a) 2633 Å
In simple words: The cut-off wavelength is the longest wavelength of light that can still cause electrons to be emitted from a metal. We find it by using the metal's work function and the formula connecting energy and wavelength.

🎯 Exam Tip: The cut-off wavelength (\( \lambda_0 \)) is related to the work function (W) by \( W = \frac{hc}{\lambda_0} \). Make sure to convert eV to Joules when using Planck's constant (h) and the speed of light (c) in SI units, or use the shortcut formula \( \lambda_0 (\text{in Å}) = \frac{12375}{W (\text{in eV})} \).

Question 20. The time taken by a photoelectron to come out after photon strikes is approximately
(a) \( 10^{-14} \) s
(b) \( 10^{-10} \) s
(c) \( 10^{-16} \) s
(d) \( 10^{-12} \) s
Answer: (b) \( 10^{-10} \) s
In simple words: The photoelectric effect happens almost instantly. The moment a photon hits the metal and has enough energy, an electron is released very, very quickly, without any noticeable delay.

🎯 Exam Tip: This instantaneous emission is a key experimental observation of the photoelectric effect, which wave theory could not explain, but particle theory (photons) could.

Question 21. A Coolidge tube operates at 37200 V, the maximum frequency of x rays emitted is
(a) 100 Hz
(b) \( 3 \times 10^{18} \) Hz
(c) \( 9 \times 10^{18} \) Hz
(d) \( 6 \times 10^{18} \) Hz
Answer: (c) \( 9 \times 10^{18} \) Hz
In simple words: In a Coolidge tube, the energy of the accelerating electrons is converted into X-ray photons. The maximum frequency of these X-rays is determined by the maximum energy the electrons gain from the operating voltage.

🎯 Exam Tip: The maximum energy of an X-ray photon (\( h\nu_{max} \)) equals the kinetic energy of the electron just before impact, which is given by \( eV \). So, \( \nu_{max} = \frac{eV}{h} \). Ensure to use consistent units for charge (e), voltage (V), and Planck's constant (h).

Question 22. De-Broglie wavelength of an electron accelerating by 100 V is
(a) 1.227 Å
(b) 122.7 µm
(c) 0.2227 m
(d) 12.27 Å
Answer: (a) 1.227 Å
In simple words: When an electron is sped up by a certain voltage, it gains kinetic energy, and because it's moving, it also has a wave-like property. We can calculate this wave's length using a special formula that links voltage to wavelength.

🎯 Exam Tip: For an electron accelerated through a potential V, the de-Broglie wavelength can be calculated using the simplified formula \( \lambda = \frac{12.27}{\sqrt{V}} \) Å, where V is in Volts.

Question 23. The energy of photon of wavelength \( \lambda \) is
(a) \( \frac{hc}{\lambda} \)
(b) \( h\lambda c \)
(c) \( \frac{\lambda}{hc} \)
(d) \( \frac{h\lambda}{c} \)
Answer: (a) \( \frac{hc}{\lambda} \)
In simple words: The energy of a tiny packet of light, called a photon, is directly related to its frequency and inversely related to its wavelength. This means shorter waves carry more energy.

🎯 Exam Tip: Remember the fundamental relationship between energy (E), Planck's constant (h), frequency (\( \nu \)), and wavelength (\( \lambda \)) for a photon: \( E = h\nu \) and since \( c = \nu\lambda \), then \( \nu = \frac{c}{\lambda} \), which gives \( E = \frac{hc}{\lambda} \).

X. Two Mark Questions:

Question 1. Why electron is preferred over X-ray in the microscope?
Answer: Electrons are preferred in microscopes over X-rays for a few reasons. Firstly, the de-Broglie wavelength of an electron is much smaller compared to that of X-rays. This is a key principle in microscopy.
\( \implies \) Because of this very small wavelength, electron microscopes can achieve a very high resolving power. This allows us to see much finer details.
\( \implies \) Therefore, we can build microscopes that reveal tiny structures, which X-rays cannot. They are essential for visualizing nanoscale objects.
In simple words: Electrons have much shorter waves than X-rays. Shorter waves let us see much smaller things, making electron microscopes more powerful to look at tiny details.

🎯 Exam Tip: The resolving power of a microscope is inversely proportional to the wavelength of the radiation used. A smaller wavelength (like that of electrons) leads to higher resolution.

Question 2. What are photoelectrons?
Answer: Photoelectrons are the electrons that are released from a metal surface. This happens when the metal is exposed to electromagnetic radiation, such as light, that has the right amount of energy and frequency.
In simple words: Photoelectrons are electrons that pop out of a metal when light shines on it with enough energy.

🎯 Exam Tip: The term "photoelectron" specifically refers to electrons emitted via the photoelectric effect, not just any free electron.

Question 3. What are matter waves?
Answer: Matter waves are a concept stating that all moving elementary particles, like electrons and protons, can also behave like waves under specific conditions. These waves show the wave-like nature of particles.
In simple words: Matter waves are the waves that moving particles, like electrons, also act like. It means tiny particles can behave as both particles and waves.

🎯 Exam Tip: The de-Broglie hypothesis introduced the idea of matter waves, stating that every moving particle has an associated wavelength given by \( \lambda = \frac{h}{p} \).

Question 4. What are continuous x rays?
Answer: Continuous X-rays produce a spectrum of radiation that includes all possible wavelengths. This range extends continuously from a certain minimum wavelength to higher values, similar to how visible light creates a continuous rainbow spectrum.
In simple words: Continuous X-rays are a type of X-ray that has a mix of all different wavelengths, not just specific ones. They are made when electrons slow down rapidly.

🎯 Exam Tip: Continuous X-rays, also known as bremsstrahlung radiation, are produced when high-speed electrons are decelerated by the electric field of atomic nuclei in the target material.

Question 5. Why is a photo-cell also called an electric eye?
Answer: A photo-cell is called an "electric eye" because it can distinguish between weak and intense light, much like a human eye. It converts light intensity into an electric current, allowing it to sense and measure light levels. This makes it useful for many automatic tasks.
In simple words: A photo-cell is like an electric eye because it can tell how bright light is and changes that light into electricity.

🎯 Exam Tip: Photocells are used in various applications like automatic doors, streetlights, and light meters due to their ability to respond to light intensity.

Question 6. What is a photoemissive cell?
Answer: A photoemissive cell is a device where the flow of electrons depends on light hitting a metal cathode. When light of a suitable frequency strikes the cathode, electrons are emitted, creating a current. This cell relies on the photoelectric effect for its operation.
In simple words: A photoemissive cell makes electricity when light hits a special metal surface, causing electrons to jump out.

🎯 Exam Tip: Photoemissive cells are characterized by their ability to emit electrons when exposed to light, typically used in applications requiring a direct conversion of light into electrical current.

Question 7. What are X-ray spectra?
Answer: X-ray spectra are patterns of X-rays produced when fast-moving electrons strike a metal target. When the intensity of these X-rays is plotted against their wavelength, it forms a curve showing both continuous and characteristic X-ray emissions. This unique pattern helps identify the target material.
In simple words: X-ray spectra are graphs that show the different types and amounts of X-rays made when electrons hit a metal, helping us see its unique X-ray "fingerprint."

🎯 Exam Tip: X-ray spectra typically consist of a continuous background (bremsstrahlung) and sharp peaks (characteristic X-rays) that are unique to the target element.

Question 8. What are photoconductive cells?
Answer: Photoconductive cells are electronic components where the resistance of a semiconductor material changes. This change happens depending on the amount of light energy that falls on it, making it more conductive when exposed to light.
In simple words: Photoconductive cells are like light-sensitive resistors; their ability to carry electricity changes when light shines on them.

🎯 Exam Tip: Photoconductive cells are often called Light Dependent Resistors (LDRs) and are used in circuits that need to detect light or measure its intensity.

Question 9. Define resolving power of the microscope.
Answer: The resolving power of a microscope refers to its ability to show two closely spaced objects as separate and distinct. This power is inversely proportional to the wavelength of the radiation used for illumination. For example, shorter wavelengths lead to better resolution.
In simple words: Resolving power is how well a microscope can show two very close things as separate instead of blurry. It gets better with shorter light waves.

🎯 Exam Tip: High resolving power is crucial for observing fine details in specimens. Electron microscopes have much higher resolving power than optical microscopes due to the extremely short de-Broglie wavelength of electrons.

Question 10. Who invented the particle nature of light?
Answer: While many scientists contributed to understanding light, Hertz's experiments confirmed that light acts as an electromagnetic wave. However, his work also showed early evidence of light behaving like particles, a concept later explained by Einstein through the photoelectric effect. This led to the understanding of light's dual particle-wave nature.
In simple words: Hertz's experiments showed that light acts like waves, but also gave hints that it could be made of particles. Einstein later explained this particle-like nature better.

🎯 Exam Tip: While Hertz observed the electromagnetic wave nature, it was primarily Max Planck (quantum hypothesis) and Albert Einstein (photoelectric effect, photon concept) who developed the particle nature of light.

Question 11. Define intensity of light.
Answer: The intensity of light defines its brightness. Scientifically, it's the amount of light energy passing through a unit area per unit time. This energy flow determines how bright or dim the light appears.
In simple words: Light intensity is simply how bright the light is, meaning how much light energy passes through a spot each second.

🎯 Exam Tip: In the photoelectric effect, light intensity is directly proportional to the number of photons, and thus to the number of emitted photoelectrons (photocurrent).

Question 12. What is diffraction?
Answer: Diffraction is the phenomenon where light waves bend around the edges of obstacles or spread out after passing through small openings. This bending causes the light to spread into regions where geometric optics predicts a shadow, creating distinctive patterns.
In simple words: Diffraction is when light waves bend around corners or spread out through tiny gaps, instead of just going straight.

🎯 Exam Tip: Diffraction is a wave property of light, evident when light passes through a slit or around an edge, and it demonstrates the wave nature of light.

Question 13. When light behaves like waves and matter?
Answer: Light exhibits a dual nature, behaving both as waves and as particles. It acts like waves when propagating through space, showing phenomena like interference and diffraction. However, during interaction with matter, such as in the photoelectric effect, it behaves like discrete particles called photons. This duality is a fundamental concept in quantum physics.
In simple words: Light acts like waves when it travels, but when it hits something, it acts like tiny particles. So, it's a wave and a particle at the same time.

🎯 Exam Tip: Understanding wave-particle duality is crucial in modern physics. Remember, light behaves as a wave during propagation and as a particle during interaction with matter.

XI. Three Marks Questions:

Question 1. What is nature of light?
Answer: Light has a dual nature, meaning it can behave both as a wave and as a particle. Its wave nature explains phenomena like interference, diffraction, and polarization. On the other hand, certain events such as black body radiation and the photoelectric effect demonstrate light's particle nature, where it acts as discrete packets of energy called photons. This duality is fundamental to how light interacts with the world.
In simple words: Light acts like a wave sometimes (like when it bends) and like tiny particles other times (like when it hits a metal). This means light has two sides to its nature.

🎯 Exam Tip: When discussing the nature of light, always refer to its wave-particle duality, providing examples for both wave-like and particle-like behaviors.

Question 2. Define one electron volt.
Answer: One electron volt (eV) is defined as the amount of kinetic energy gained by an electron. This occurs when the electron is accelerated through an electric potential difference of exactly one volt. It's a convenient unit for energy in atomic and particle physics.
\( \implies \) Mathematically, 1 eV is equal to the product of the elementary charge (q) and the potential difference (V).
\( \implies \) So, \( 1 \text{ eV} = (1.602 \times 10^{-19} \text{ C}) \times (1 \text{ V}) \).
\( \implies \) This calculation results in \( 1 \text{ eV} = 1.602 \times 10^{-19} \) Joules.
In simple words: One electron volt is the energy an electron gets when it speeds up through a 1-volt electric field. It's a tiny amount of energy.

🎯 Exam Tip: Clearly state the definition and the conversion factor between electron volts and Joules. This conversion is frequently needed in calculations.

Question 3. Differentiate particle and wave.
Answer: A particle is considered a tiny, localized concentration of matter, existing in a specific spot in space and time. It has properties like definite mass, charge, and momentum. In contrast, a wave is a broad distribution of energy that is not localized. It spreads out over space and time, characterized by properties like wavelength, frequency, and amplitude. Waves transmit energy without transferring matter.
In simple words: A particle is a small, solid thing in one place, like a tiny ball. A wave is energy that spreads out, like ripples in water, and doesn't stay in one spot.

🎯 Exam Tip: When differentiating, focus on key characteristics like localization (particle is localized, wave is distributed), energy transfer (particle carries energy with matter, wave carries energy without matter), and associated properties (mass for particle, wavelength/frequency for wave).

XII. Five Mark Questions:

Question 1. What are the applications of the photocells?
Answer: Photocells have a wide range of practical applications, especially in modern technology, due to their light-sensing capabilities:
1. They are used in automatic switches and sensors, for example, to turn on lights when it gets dark.
2. Photocells enable automatic switching ON and OFF of ordinary lights and streetlights, saving energy.
3. They play a crucial role in motion picture reproduction, helping to convert light signals back into sound.
4. In sports, photocells are utilized as timers to accurately measure the speeds of athletes during races, providing precise results.
5. They are also used to measure the intensity of a given light source and calculate the exact time of exposure in photography and other optical instruments.
In simple words: Photocells are used in many ways, like turning on streetlights when it gets dark, helping movie sound work, timing races, and measuring how bright light is.

🎯 Exam Tip: When listing applications, aim for variety and clarity. Briefly explain how photocells function in each application to demonstrate a deeper understanding.

Question 2. Write down the characteristics of photons.
Answer: According to the particle nature of light, photons possess several distinct characteristics:
* Photons are elementary particles of light with a frequency \( \nu \) and wavelength \( \lambda \). Each photon has energy given by \( E = h\nu = \frac{hc}{\lambda} \).
* The energy of a single photon is determined solely by its frequency, not by the intensity of the light source. The intensity of light relates to the number of photons, not their individual energy.
* Photons travel at the speed of light (c) in a vacuum, which is approximately \( 3 \times 10^8 \) m/s, and they carry momentum given by \( p = \frac{h}{\lambda} \).
* Photons have no electric charge, meaning they are electrically neutral. As a result, they are unaffected by electric or magnetic fields.
* When a photon interacts with matter, such as in a photon-electron collision, the total energy, total linear momentum, and total angular momentum are conserved. However, the number of photons may not be conserved in such interactions, as photons can be absorbed or new ones produced.
In simple words: Photons are tiny light particles that have energy based on their color (frequency) and travel at light speed. They have no electric charge and carry momentum. When they hit things, energy is conserved, but the number of photons can change.

🎯 Exam Tip: Clearly list each characteristic of a photon, emphasizing its energy-frequency relationship, lack of charge, constant speed, and role in light-matter interactions. These points are central to understanding quantum optics.

Question 3. Explain characteristic X-rays and their applications in medical therapy and industry.
Answer:
Characteristic X-ray Spectra:
Characteristic X-rays produce narrow, well-defined peaks at specific wavelengths when a target material is bombarded with fast electrons. This line spectrum is unique to each element.
1. The process begins when an energetic electron removes an electron from an inner shell (like the K-shell) of a target atom, creating a vacancy.
2. To fill this vacancy, an electron from an outer orbit (e.g., L, M, or N shell) jumps down to the lower energy shell.
3. During this transition, the energy difference between the two shells is released as an X-ray photon with a definite and characteristic wavelength. This energy is unique to the atom.

The following diagram illustrates the origin of characteristic X-ray spectra:

🎯 Exam Tip: For characteristic X-rays, remember that their wavelengths are specific to the atomic number of the target material, unlike continuous X-rays. Focus on outlining distinct applications in both medical and industrial fields.

XIII. Conceptual Questions:

Question 1. Are other particles, other than electrons having a wave nature?
Answer: Yes, the wave nature is not exclusive to electrons; other particles also exhibit wave-like properties. Particles such as neutrons and alpha particles are associated with waves. This can be observed through experiments where these particles undergo diffraction when they are scattered by suitable crystals, just like waves. This observation confirms the de-Broglie hypothesis, which states that all matter possesses wave-like characteristics.
In simple words: Yes, not just electrons, but other tiny particles like neutrons and alpha particles also act like waves. We can see them bend and spread out when they pass through crystals.

🎯 Exam Tip: Emphasize that wave-particle duality applies to all matter, not just photons or electrons. Diffraction experiments with neutrons and protons provide strong evidence for this concept.

Question 2. Why diffraction effects of ordinary light is very small?
Answer: The diffraction effects of ordinary light are usually very small and not easily noticeable because the wavelength of visible light is much larger compared to the spacing between atoms in crystal lattice planes. Significant diffraction occurs when the wavelength of the incident radiation is comparable to the size of the obstacle or the spacing of the grating. For visible light, typical obstacles are much larger than its wavelength, leading to minimal bending.
In simple words: We don't usually see light bending much because light waves are too long compared to most things they bend around. To see clear bending, the object or hole needs to be as small as the light's wavelength.

🎯 Exam Tip: Remember that observable diffraction requires the wavelength to be comparable to or larger than the diffracting object or aperture. This explains why X-rays are used to study crystal structures (because their wavelengths match inter-atomic spacing).

Question 3. Why crystals are used for three-dimensional grating?
Answer: Crystals are used as three-dimensional gratings because their atomic spacing is very regular and on the order of \( 10^{-10} \) meters. This spacing is comparable to the wavelengths of X-rays and matter waves (like electron beams). When X-rays or electron beams pass through a crystal, they are diffracted by the orderly arrangement of atoms, forming a distinct pattern. This pattern acts like a fingerprint, allowing scientists to study the internal structure of the crystal.
In simple words: Crystals are used like special 3D sieves because the gaps between their atoms are just the right size to bend X-rays and electron waves. This helps us see how the atoms are arranged inside.

🎯 Exam Tip: The key reason crystals act as excellent diffraction gratings is the match between their inter-atomic spacing and the wavelengths of X-rays or matter waves. This principle is fundamental to X-ray crystallography and electron diffraction.

XIV. Additional Problems:

Question 1. Calculate the momentum and the de- Broglie wavelength of an electron with kinetic energy 25 eV.
Answer: We need to calculate two things for an electron with a given kinetic energy: its momentum and its de Broglie wavelength. First, we convert the kinetic energy from electron Volts (eV) to Joules (J).

Given:
Kinetic Energy \( K.E. = 25 \text{ eV} \)
Mass of electron \( m = 9.1 \times 10^{-31} \text{ kg} \)
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)
Charge of electron \( e = 1.6 \times 10^{-19} \text{ C} \)

**Convert K.E. to Joules:**
\( K.E. = 25 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 4.0 \times 10^{-18} \text{ J} \)

**(i) Momentum of the electron (p):**
The momentum of an electron can be found using its kinetic energy. We use the formula \( K.E. = \frac{p^2}{2m} \), which can be rearranged to \( p = \sqrt{2mK.E.} \). This momentum is tiny, showing how light electrons are and how quantum effects become important at their scale.
\( p = \sqrt{2 \times 9.1 \times 10^{-31} \text{ kg} \times 4.0 \times 10^{-18} \text{ J}} \)
\( p = \sqrt{72.8 \times 10^{-49}} \)
\( p = \sqrt{7.28 \times 10^{-48}} \)
\( p = 2.698 \times 10^{-24} \text{ kg ms}^{-1} \)

**(ii) De Broglie wavelength (\( \lambda \)):**
Next, we calculate the de Broglie wavelength (\( \lambda \)) using Planck's constant (h) and the electron's momentum (p). The de Broglie wavelength shows the wave-like nature of particles.
\( \lambda = \frac{h}{p} \)
\( \lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{2.698 \times 10^{-24} \text{ kg ms}^{-1}} \)
\( \lambda = 2.455 \times 10^{-10} \text{ m} \)
\( \lambda = 2.455 \text{ Å} \)
In simple words: We first converted the electron's energy to Joules. Then, we used formulas to find its momentum and its wave-like length, called the de Broglie wavelength.

🎯 Exam Tip: Remember to convert all energy values to Joules before using them in formulas involving Planck's constant to avoid calculation errors.

Question 2. Monochromatic light of frequency 6 x 10¹⁴ Hz is produced by a laser. The power emitted is 2 x 10^-3 W.
(i) What is the energy of each photon in the light?
(ii) How many photons per second, on average, are emitted by the source?
Answer: We need to determine the energy of a single photon and the total number of photons emitted per second by a laser. We will use Planck's formula to find the energy per photon and then relate it to the laser's power output.

**Given:**
Frequency \( \nu = 6 \times 10^{14} \text{ Hz} \)
Power emitted \( P = 2 \times 10^{-3} \text{ W} \)
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)

**(i) Energy of each photon (E):**
To find the energy of each photon, we use Planck's formula \( E = h\nu \). Each photon carries a very small amount of energy, which is a fundamental concept in quantum physics.
\( E = 6.626 \times 10^{-34} \text{ Js} \times 6 \times 10^{14} \text{ Hz} \)
\( E = 3.9756 \times 10^{-19} \text{ J} \)
\( E \approx 3.98 \times 10^{-19} \text{ J} \)

**(ii) Number of photons per second (N):**
To find the number of photons emitted per second, we divide the total power emitted by the laser by the energy of a single photon. Power is the total energy released per second. Lasers produce a huge number of photons every second, which is why they can deliver a lot of energy.
\( P = N \times E \)
\( N = \frac{P}{E} \)
\( N = \frac{2 \times 10^{-3} \text{ W}}{3.98 \times 10^{-19} \text{ J}} \)
\( N = 5.025 \times 10^{15} \text{ photons/second} \)
\( N \approx 5 \times 10^{15} \text{ photons/second} \)
In simple words: We calculated how much energy one light particle has based on its color. Then, we used the laser's total power to figure out how many such light particles it shoots out every second.

🎯 Exam Tip: Remember that power is energy per unit time, so dividing total power by the energy of one photon gives the number of photons per second.

Question 3. A proton is moving at a speed of 0.900 times the velocity of light. Find the kinetic energy in Joules and Mev.
Answer: The question asks for the kinetic energy of a proton moving at a very high speed, both in Joules and Mega-electron Volts (MeV). We will use the classical kinetic energy formula and then convert between units.

**Given:**
Speed of light \( c = 3 \times 10^8 \text{ m/s} \)
Speed of proton \( v = 0.900 \times c = 0.900 \times 3 \times 10^8 \text{ m/s} = 2.7 \times 10^8 \text{ m/s} \)
Mass of proton \( m = 1.673 \times 10^{-27} \text{ kg} \)
(Note: Relativistic effects are significant at this speed, but we follow the classical formula implied by the source steps.)

**Kinetic Energy (K.E.) in Joules:**
We use the classical kinetic energy formula \( K.E. = \frac{1}{2} mv^2 \). This value represents the energy of motion for the proton.
\( K.E. = \frac{1}{2} \times 1.673 \times 10^{-27} \text{ kg} \times (2.7 \times 10^8 \text{ m/s})^2 \)
\( K.E. = \frac{1}{2} \times 1.673 \times 10^{-27} \times 7.29 \times 10^{16} \text{ J} \)
\( K.E. = 6.098 \times 10^{-11} \text{ J} \)

**Kinetic Energy (K.E.) in MeV:**
To convert Joules to electron Volts (eV), we divide by the charge of an electron (\( 1.6 \times 10^{-19} \text{ J/eV} \)). Then, to convert eV to MeV (Mega-electron Volts), we divide by \( 10^6 \). It shows that even a tiny proton can carry substantial energy at high speeds.
\( K.E. = \frac{6.098 \times 10^{-11} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 3.811 \times 10^8 \text{ eV} \)
\( K.E. = \frac{3.811 \times 10^8 \text{ eV}}{10^6 \text{ eV/MeV}} = 381.1 \text{ MeV} \)
In simple words: We calculated how much energy a very fast proton has, first in Joules and then converted it to Mega-electron Volts, which is a unit often used for tiny particles.

🎯 Exam Tip: Pay close attention to unit conversions, especially between Joules, electron Volts, and Mega-electron Volts, as these are common sources of error.

Question 4. Calculate the momentum of an electron with kinetic energy 2 eV.
Answer: To calculate the momentum of an electron, we first need to convert its kinetic energy from electron Volts (eV) to Joules (J). We then use the kinetic energy formula to find the momentum.

**Given:**
Kinetic Energy \( K.E. = 2 \text{ eV} \)
Mass of electron \( m = 9.1 \times 10^{-31} \text{ kg} \)
Charge of electron \( e = 1.6 \times 10^{-19} \text{ C} \)

**Convert K.E. to Joules:**
\( K.E. = 2 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 3.2 \times 10^{-19} \text{ J} \)

**Momentum (p):**
We use the formula \( K.E. = \frac{p^2}{2m} \), which means \( p = \sqrt{2mK.E.} \). The momentum of subatomic particles like electrons is very small, which is why their wave nature is usually not noticeable in everyday life.
\( p = \sqrt{2 \times 9.1 \times 10^{-31} \text{ kg} \times 3.2 \times 10^{-19} \text{ J}} \)
\( p = \sqrt{58.24 \times 10^{-50}} \)
\( p = 7.6315 \times 10^{-25} \text{ kg ms}^{-1} \)
\( p \approx 7.63 \times 10^{-25} \text{ kg ms}^{-1} \)
In simple words: We changed the electron's energy from electron Volts to Joules. Then, using a formula that connects energy and momentum, we found how much "push" the electron has.

🎯 Exam Tip: Always remember the conversion factor for electron Volts to Joules (\( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \)) and use it correctly.

Question 5. How many photons of frequency 10¹⁴ Hz will make up 19.86 J of energy?
Answer: We need to find out how many photons are required to make up a specific total energy, given the frequency of each photon. First, we calculate the energy of a single photon.

**Given:**
Total Energy \( E_{total} = 19.86 \text{ J} \)
Frequency \( \nu = 10^{14} \text{ Hz} \)
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)

**Energy of one photon (E):**
Using the formula \( E = h\nu \), we can find the energy carried by one photon. Each photon carries a discrete packet of energy.
\( E = 6.626 \times 10^{-34} \text{ Js} \times 10^{14} \text{ Hz} \)
\( E = 6.626 \times 10^{-20} \text{ J} \)

**Number of photons (n):**
To find the total number of photons, we divide the total energy by the energy of one photon. Even small amounts of energy can correspond to an incredibly large number of photons, highlighting the discrete nature of light.
\( n = \frac{E_{total}}{E} \)
\( n = \frac{19.86 \text{ J}}{6.626 \times 10^{-20} \text{ J}} \)
\( n = 2.997 \times 10^{20} \)
\( n \approx 3 \times 10^{20} \text{ photons} \)
In simple words: We calculated the energy of one light particle using its frequency. Then, we divided the total energy by this single particle's energy to find out how many light particles are needed in total.

🎯 Exam Tip: This type of problem often tests your understanding of the quantization of energy, where light comes in discrete packets (photons).

Question 6. What is the (a) momentum, (b) speed, and (c) de-Broglie wavelength of an electron with the kinetic energy of 120 eV.
Answer: We need to calculate the momentum, speed, and de Broglie wavelength of an electron given its kinetic energy. We start by converting the kinetic energy to Joules.

**Given:**
Kinetic energy \( K.E. = 120 \text{ eV} \)
Mass of electron \( m = 9.1 \times 10^{-31} \text{ kg} \)
Planck's constant \( h = 6.626 \times 10^{-34} \text{ Js} \)
Charge of electron \( e = 1.6 \times 10^{-19} \text{ C} \)

**Convert K.E. to Joules:**
\( K.E. = 120 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.92 \times 10^{-17} \text{ J} \)

**(a) Momentum (P) of an electron:**
Using the formula \( K.E. = \frac{P^2}{2m} \), we can find the momentum \( P = \sqrt{2mK.E.} \). This tells us the electron's linear motion.
\( P = \sqrt{2 \times 9.1 \times 10^{-31} \text{ kg} \times 1.92 \times 10^{-17} \text{ J}} \)
\( P = \sqrt{34.944 \times 10^{-48}} \)
\( P = 5.911 \times 10^{-24} \text{ kg ms}^{-1} \)
\( P \approx 5.91 \times 10^{-24} \text{ kg ms}^{-1} \)

**(b) Speed (v) of an electron:**
From the definition of momentum \( P = mv \), we can find the speed \( v = \frac{P}{m} \). This represents how fast the electron is moving.
\( v = \frac{5.911 \times 10^{-24} \text{ kg ms}^{-1}}{9.1 \times 10^{-31} \text{ kg}} \)
\( v = 0.6495 \times 10^7 \text{ m/s} \)
\( v \approx 6.5 \times 10^6 \text{ m/s} \)

**(c) De Broglie wavelength (\( \lambda \)):**
Using the de Broglie wavelength formula \( \lambda = \frac{h}{P} \), we can find its wave-like property. The small wavelength of electrons is what makes electron microscopes so powerful, allowing them to image objects much smaller than what visible light can resolve.
\( \lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{5.911 \times 10^{-24} \text{ kg ms}^{-1}} \)
\( \lambda = 1.121 \times 10^{-10} \text{ m} \)
\( \lambda = 0.1121 \text{ nm} \)
\( \lambda \approx 0.112 \text{ nm} \)
In simple words: First, we converted the electron's energy. Then, we found its momentum, its speed, and finally, its wave-like length using specific physics formulas.

🎯 Exam Tip: When working with electrons, remember their tiny mass and the appropriate Planck's constant value to ensure accurate calculations for quantum properties.