JEE Physics Atoms MCQs Set 13

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Chapter-wise Objective Questions: Atoms

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Multiple Choice Questions

Question 1: The de Broglie wavelength of an electron in the first Bohr orbit of hydrogen is nearly (Bohr radius 0.53 Å):
(a) 1.7 Å
(b) 3.3 Å
(c) 0.53 Å
(d) 6.6 Å
Show Answer & Explanation

Answer: (b) 3.3 Å

Explanation:
1. Bohr's condition \( 2\pi r_n=n\lambda \) gives for \( n=1 \): \( \lambda=2\pi r_1=2\pi\times0.53=3.3\text{ Å} \).

Teacher's Note:
1. The first orbit holds exactly one de Broglie wavelength.
2. The wavelength equals the circumference of the orbit.

Question 2: A positive hydrogen-like ion with its electron in the ground state ejects it when a photon of wavelength 228 Å or less is absorbed. The ion is:
(a) \( \text{Li}^{2+} \)
(b) \( \text{Be}^{3+} \)
(c) \( \text{H} \)
(d) \( \text{He}^+ \)
Show Answer & Explanation

Answer: (d) \( \text{He}^+ \)

Explanation:
1. Ionisation energy \( =\dfrac{12400}{228}=54.4\text{ eV} \).
2. Since \( 13.6Z^2=54.4 \), \( Z=2 \): the ion is \( \text{He}^+ \).

Teacher's Note:
1. Ionisation energy is Z squared times 13.6 eV.
2. The threshold wavelength is the longest wavelength that can ionise.

Question 3: The temperature at which the average kinetic energy of hydrogen molecules, taken as \( \tfrac32kT \), equals the binding energy of the electron in the ground state of hydrogen (13.6 eV) is nearly (\( k=1.38\times10^{-23}\text{ J/K} \)):
(a) \( 1.05\times10^4\text{ K} \)
(b) \( 2.1\times10^5\text{ K} \)
(c) \( 1.05\times10^5\text{ K} \)
(d) \( 3.5\times10^5\text{ K} \)
Show Answer & Explanation

Answer: (c) \( 1.05\times10^5\text{ K} \)

Explanation:
1. \( T=\dfrac{2E}{3k}=\dfrac{2\times13.6\times1.6\times10^{-19}}{3\times1.38\times10^{-23}}\approx1.05\times10^5\text{ K} \).

Teacher's Note:
1. Convert the energy to joules.
2. This temperature shows why hydrogen is ionised in stars.

Question 4: Find the smallest wavelength in the emission spectrum of (a) hydrogen and (b) He\( ^+ \). The answers are nearly:
(a) 912 Å and 228 Å
(b) 1215 Å and 304 Å
(c) 912 Å and 456 Å
(d) 228 Å and 912 Å
Show Answer & Explanation

Answer: (a) 912 Å and 228 Å

Explanation:
1. The smallest wavelength is the series limit of the Lyman series, \( \dfrac1\lambda=RZ^2 \).
2. For hydrogen \( \lambda=\dfrac{12400}{13.6}=912\text{ Å} \); for \( \text{He}^+ \) \( \lambda=\dfrac{912}{4}=228\text{ Å} \).

Teacher's Note:
1. The most energetic photon corresponds to a transition from infinity to the ground state.
2. Wavelength scales as one over Z squared.

Question 5: An electron is in the second Bohr orbit of a He\( ^+ \) ion. Its angular frequency of revolution is nearly:
(a) \( 4.1\times10^{16}\text{ rad/s} \)
(b) \( 1.0\times10^{16}\text{ rad/s} \)
(c) \( 2.1\times10^{15}\text{ rad/s} \)
(d) \( 2.1\times10^{16}\text{ rad/s} \)
Show Answer & Explanation

Answer: (d) \( 2.1\times10^{16}\text{ rad/s} \)

Explanation:
1. Speed \( v=2.19\times10^6\dfrac{Z}{n}=2.19\times10^6\text{ m/s} \) and radius \( r=\dfrac{n^2a_0}{Z}=1.06\times10^{-10}\text{ m} \).
2. \( \omega=\dfrac vr=\dfrac{2.19\times10^6}{1.06\times10^{-10}}=2.1\times10^{16}\text{ rad/s} \).

Teacher's Note:
1. Both the speed and the radius have simple Z and n dependence in Bohr's model.
2. omega is proportional to Z squared over n cubed.

Question 6: A stationary hydrogen atom emits a photon corresponding to the first line of the Lyman series. The velocity acquired by the atom is nearly (mass of hydrogen atom \( 1.67\times10^{-27}\text{ kg} \)):
(a) 33 m/s
(b) 3.3 m/s
(c) 0.33 m/s
(d) 3.3 km/s
Show Answer & Explanation

Answer: (b) 3.3 m/s

Explanation:
1. Photon energy \( 10.2\text{ eV} \) and momentum \( \dfrac Ec=\dfrac{10.2\times1.6\times10^{-19}}{3\times10^8}=5.4\times10^{-27}\text{ kg m/s} \).
2. By momentum conservation \( v=\dfrac{p}{m}=\dfrac{5.4\times10^{-27}}{1.67\times10^{-27}}=3.3\text{ m/s} \).

Teacher's Note:
1. The recoil momentum equals the photon momentum.
2. Recoil energy is a tiny fraction of the photon energy.

Question 7: For a gas of Li\( ^{2+} \) ions, the wavelength of radiation required to excite the electron from \( n=1 \) to \( n=3 \), and the number of spectral lines in the emission spectrum of the excited system, are nearly:
(a) 114 Å and 3 lines
(b) 114 Å and 6 lines
(c) 228 Å and 3 lines
(d) 912 Å and 3 lines
Show Answer & Explanation

Answer: (a) 114 Å and 3 lines

Explanation:
1. Energy required \( 13.6\times9\left(1-\dfrac19\right)=108.8\text{ eV} \), so \( \lambda=\dfrac{12400}{108.8}=114\text{ Å} \).
2. From \( n=3 \) the possible transitions are 3 to 2, 3 to 1 and 2 to 1: three lines.

Teacher's Note:
1. The number of lines from level n is n(n-1)/2.
2. Use Z = 3 for lithium.

Question 8: A hydrogen atom moving with kinetic energy \( K \) collides head-on and inelastically with another stationary hydrogen atom, both in the ground state, so that one of them emits a photon. The minimum value of \( K \) is:
(a) 10.2 eV
(b) 13.6 eV
(c) 20.4 eV
(d) 27.2 eV
Show Answer & Explanation

Answer: (c) 20.4 eV

Explanation:
1. In a perfectly inelastic head-on collision of equal masses, half of the kinetic energy can be lost: \( \Delta K=\dfrac K2 \).
2. The smallest excitation energy is 10.2 eV, so \( \dfrac K2=10.2 \) and \( K=20.4\text{ eV} \).

Teacher's Note:
1. The kinetic energy of the centre of mass cannot be used for excitation.
2. The reduced mass gives half of the energy for equal masses.

Question 9: The cut-off wavelength for the continuous X-rays from a tube operating at 40 kV is, and the lowest three wavelengths from a tube at 20 kV where an electron converts 70% of its energy into a photon at each collision are:
(a) 0.62 Å; 0.89 Å, 2.95 Å, 9.84 Å
(b) 0.31 Å; 0.89 Å, 2.95 Å, 9.84 Å
(c) 0.31 Å; 0.62 Å, 2.95 Å, 9.84 Å
(d) 0.31 Å; 0.89 Å, 1.77 Å, 4.4 Å
Show Answer & Explanation

Answer: (b) 0.31 Å; 0.89 Å, 2.95 Å, 9.84 Å

Explanation:
1. Cut-off wavelength \( \lambda_{min}=\dfrac{12400}{40000}=0.31\text{ Å} \).
2. At 20 kV: first photon \( 0.7\times20=14\text{ keV} \) (\( 0.89\text{ Å} \)); the electron keeps 6 keV, so the second photon is \( 0.7\times6=4.2\text{ keV} \) (\( 2.95\text{ Å} \)); it keeps 1.8 keV and the third is \( 1.26\text{ keV} \) (\( 9.84\text{ Å} \)).

Teacher's Note:
1. Each collision takes 70% of the remaining energy.
2. Wavelength in Å is 12.4 divided by the energy in keV.

Question 10: The wavelength of the \( K_\alpha \) line in iron (Z = 26) is 193 pm. Taking the screening constant as 1, the wavelength of the \( K_\alpha \) line in copper (Z = 29) is nearly:
(a) 173 pm
(b) 215 pm
(c) 193 pm
(d) 154 pm
Show Answer & Explanation

Answer: (d) 154 pm

Explanation:
1. Moseley's law: \( \dfrac1\lambda\propto(Z-1)^2 \).
2. \( \lambda_{Cu}=193\left(\dfrac{25}{28}\right)^2=154\text{ pm} \).

Teacher's Note:
1. Subtract 1 from each atomic number.
2. A heavier element has a shorter wavelength.

Question 11: A hydrogen-like atom of atomic number \( Z \) is in a high excited state of quantum number \( n \). It can reach the first excited state by emitting two photons of energies 10.20 eV and 17.00 eV in succession, or the second excited state by emitting 4.25 eV and 5.95 eV in succession. The values of \( n \) and \( Z \) are:
(a) n = 5, Z = 3
(b) n = 6, Z = 2
(c) n = 6, Z = 3
(d) n = 4, Z = 3
Show Answer & Explanation

Answer: (c) n = 6, Z = 3

Explanation:
1. From \( n \) to 2: \( 27.2\text{ eV} \); from \( n \) to 3: \( 10.2\text{ eV} \). The difference is \( E_3-E_2=17.0\text{ eV} \).
2. \( 13.6Z^2\left(\dfrac14-\dfrac19\right)=17\Rightarrow Z^2=9 \), so \( Z=3 \).
3. Then \( 122.4\left(\dfrac14-\dfrac1{n^2}\right)=27.2\Rightarrow n=6 \).

Teacher's Note:
1. Subtracting the two energy totals isolates the 3-to-2 transition.
2. Use Z-squared times 13.6 eV for the energy scale.

Question 12: An electron in the hydrogen atom makes transitions as the total energy changes. The energy required to remove both electrons from a neutral helium atom is, given that 24.6 eV is needed to remove one electron:
(a) 79.0 eV
(b) 49.2 eV
(c) 38.2 eV
(d) 51.8 eV
Show Answer & Explanation

Answer: (a) 79.0 eV

Explanation:
1. Removing the second electron means ionising He\( ^+ \): \( 13.6\times Z^2=13.6\times4=54.4\text{ eV} \).
2. Total \( 24.6+54.4=79.0\text{ eV} \).

Teacher's Note:
1. The first and second ionisation energies add.
2. The second electron sees Z = 2 with a single electron.

Question 13: The wavelength of the first line of the Balmer series of hydrogen is 6561 Å. The wavelength of the second spectral line of the Balmer series of singly ionised helium is:
(a) 1640 Å
(b) 2430 Å
(c) 4687 Å
(d) 1215 Å
Show Answer & Explanation

Answer: (d) 1215 Å

Explanation:
1. Hydrogen first Balmer line (3 to 2): \( \dfrac1{\lambda_H}=R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{36} \).
2. He\( ^+ \) second line of its Balmer series (4 to 2): \( \dfrac1{\lambda}=4R\left(\dfrac14-\dfrac1{16}\right)=\dfrac{3R}{4} \).
3. Ratio \( \dfrac{\lambda}{\lambda_H}=\dfrac{5/36}{3/4}=\dfrac5{27} \), so \( \lambda=6561\times\dfrac5{27}=1215\text{ Å} \).

Teacher's Note:
1. The Balmer series ends at n = 2 for any hydrogen-like ion.
2. The factor Z squared multiplies the Rydberg term.

Question 14: An X-ray photon of wavelength \( \lambda \) and frequency \( \nu \) collides with a stationary free electron and bounces off. If \( \lambda' \) and \( \nu' \) are the wavelength and frequency of the scattered photon, then:
(a) \( \lambda'=\lambda,\ \nu'=\nu \)
(b) \( \lambda'>\lambda,\ \nu'<\nu \)
(c) \( \lambda'<\lambda,\ \nu'>\nu \)
(d) \( \lambda'>\lambda,\ \nu'>\nu \)
Show Answer & Explanation

Answer: (b) \( \lambda'>\lambda,\ \nu'<\nu \)

Explanation:
1. The electron recoils and gains kinetic energy, taken from the photon.
2. The photon therefore has less energy: lower frequency and longer wavelength (Compton effect).

Teacher's Note:
1. Energy and momentum are conserved in the collision.
2. The wavelength shift is \( \Delta\lambda=\dfrac{h}{mc}(1-\cos\theta) \).

Question 15: The K X-ray wavelengths of two metals A and B are \( \dfrac{4}{1875R} \) and \( \dfrac{1}{675R} \) respectively, where \( R \) is the Rydberg constant. The number of elements lying between A and B in the periodic table is:
(a) 4
(b) 3
(c) 5
(d) 6
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. \( \dfrac1\lambda=\tfrac34R(Z-1)^2 \). For A: \( \dfrac{1875R}{4}=\dfrac{3R}{4}(Z-1)^2\Rightarrow(Z-1)^2=625 \), so \( Z=26 \).
2. For B: \( 675R=\dfrac{3R}{4}(Z-1)^2\Rightarrow(Z-1)^2=900 \), so \( Z=31 \).
3. Elements between: 27, 28, 29, 30, that is 4.

Teacher's Note:
1. Use the K-alpha Moseley relation with screening constant 1.
2. Count only the elements strictly between the two.

Question 16: The photoelectric threshold wavelength of tungsten is 230 nm. The energy of electrons ejected by ultraviolet light of wavelength 180 nm is nearly:
(a) 0.15 eV
(b) 15 eV
(c) 1.5 eV
(d) 1.5 keV
Show Answer & Explanation

Answer: (c) 1.5 eV

Explanation:
1. \( K=1240\left(\dfrac1{180}-\dfrac1{230}\right)=6.89-5.39=1.5\text{ eV} \).

Teacher's Note:
1. The kinetic energy is the photon energy minus the work function.
2. The work function is hc divided by the threshold wavelength.

Question 17: In an X-ray tube the electrons strike the target with a speed that is 10% of the speed of light. The applied voltage is nearly:
(a) 1.05 kV
(b) 2.56 kV
(c) 5.68 kV
(d) 517 V
Show Answer & Explanation

Answer: (b) 2.56 kV

Explanation:
1. \( eV=\tfrac12mv^2=\tfrac12\times9.1\times10^{-31}\times(3\times10^7)^2=4.1\times10^{-16}\text{ J} \).
2. \( V=\dfrac{4.1\times10^{-16}}{1.6\times10^{-19}}\approx2.56\text{ kV} \).

Teacher's Note:
1. At 10% of c the non-relativistic formula is accurate to better than 1%.
2. The potential difference equals kinetic energy in eV.

Question 18: Which of the following are correct for a hypothetical hydrogen-like atom whose spectral lines for transitions from \( n=p \) to \( n=1 \) have wavelengths \( \lambda=\dfrac{1500p^2}{p^2-1}\text{ Å} \) (take \( hc=12400\text{ eV Å} \))? (1) The longest and shortest wavelengths in the series are 2000 Å and 1500 Å. (2) The energy difference between the fourth and third orbits is about 0.40 eV. (3) The energy of the second orbit is 6.2 eV. (4) The ionisation potential is about 8.27 V.
(a) (1), (3) and (4) only
(b) (2), (3) and (4) only
(c) (1), (2) and (3) only
(d) (1), (2) and (4) only
Show Answer & Explanation

Answer: (d) (1), (2) and (4) only

Explanation:
1. Longest wavelength is \( p=2 \): \( \dfrac{1500\times4}{3}=2000\text{ Å} \); shortest is as \( p\to\infty \): 1500 Å. So (1) is true.
2. The photon energy limit is \( \dfrac{12400}{1500}=8.27\text{ eV} \), the ionisation energy (4) is true; \( E_n=-\dfrac{8.27}{n^2} \).
3. \( E_4-E_3=8.27\left(\dfrac19-\dfrac1{16}\right)=0.40\text{ eV} \) (2) is true, while \( E_2=-2.07\text{ eV} \) so (3) is false.

Teacher's Note:
1. Work out the energy levels from the series limit.
2. The second orbit energy is negative and smaller in magnitude than 6.2 eV.

Question 19: A hydrogen atom in its ground state is excited by monochromatic radiation of wavelength 975 Å (ionisation energy 13.6 eV). Which statements are correct? (1) The emission spectrum has 6 lines. (2) The energy difference between the third and fourth orbits is 0.66 eV. (3) The longest wavelength emitted is about 1.875 \( \mu \)m. (4) The smallest wavelength emitted is 975 Å.
(a) (1), (2) and (3) only
(b) (2), (3) and (4) only
(c) (1), (2), (3) and (4)
(d) (1) and (4) only
Show Answer & Explanation

Answer: (c) (1), (2), (3) and (4)

Explanation:
1. Photon energy \( \dfrac{12400}{975}=12.7\text{ eV} \) raises the atom to \( n=4 \) (\( 13.6\left(1-\dfrac1{16}\right)=12.75\text{ eV} \)).
2. From \( n=4 \) there are \( \dfrac{4\times3}{2}=6 \) lines. \( E_4-E_3=13.6\left(\dfrac19-\dfrac1{16}\right)=0.66\text{ eV} \).
3. Longest wavelength is 4 to 3: \( \dfrac{12400}{0.66}=18800\text{ Å}=1.88\ \mu\text{m} \); the shortest is 4 to 1, which is the 975 Å photon. All four are correct.

Teacher's Note:
1. Count the lines using n(n-1)/2.
2. The transition 4 to 1 has the same energy as the absorbed photon.

Question 20: In a hydrogen-like atom an electron makes a transition from level \( n \) to \( n-1 \). For \( n\gg1 \) the frequency of the radiation emitted is proportional to:
(a) \( \dfrac1{n^3} \)
(b) \( \dfrac1{n^2} \)
(c) \( n^2 \)
(d) \( \dfrac1{n^4} \)
Show Answer & Explanation

Answer: (a) \( \dfrac1{n^3} \)

Explanation:
1. \( \nu\propto\dfrac1{(n-1)^2}-\dfrac1{n^2}=\dfrac{2n-1}{n^2(n-1)^2}\approx\dfrac{2}{n^3} \) for large \( n \).

Teacher's Note:
1. For large n the frequency equals the orbital frequency, as the correspondence principle requires.
2. Orbital frequency goes as one over n cubed.

Question 21: Ultraviolet light of two wavelengths \( \lambda_1 \) and \( \lambda_2 \) falls on hydrogen atoms in the ground state and liberates electrons with kinetic energies 1.8 eV and 4.0 eV respectively. The ratio \( \lambda_1/\lambda_2 \) is:
(a) \( \dfrac78 \)
(b) \( \dfrac9{20} \)
(c) \( \dfrac{20}{9} \)
(d) \( \dfrac87 \)
Show Answer & Explanation

Answer: (d) \( \dfrac87 \)

Explanation:
1. Photon energies: \( 13.6+1.8=15.4\text{ eV} \) and \( 13.6+4.0=17.6\text{ eV} \).
2. \( \dfrac{\lambda_1}{\lambda_2}=\dfrac{E_2}{E_1}=\dfrac{17.6}{15.4}=\dfrac87 \).

Teacher's Note:
1. The ionisation energy of hydrogen is 13.6 eV.
2. Wavelength is inversely proportional to photon energy.

Question 22: A 5 MeV alpha particle is scattered through 180 degrees by a fixed uranium nucleus (Z = 92). The distance of closest approach is of the order of:
(a) \( 10^{-15}\text{ cm} \)
(b) \( 10^{-10}\text{ cm} \)
(c) \( 10^{-12}\text{ cm} \)
(d) \( 1\text{ Å} \)
Show Answer & Explanation

Answer: (c) \( 10^{-12}\text{ cm} \)

Explanation:
1. \( K=\dfrac{1}{4\pi\epsilon_0}\dfrac{(2e)(92e)}{r_0} \), so \( r_0=\dfrac{9\times10^9\times184\times(1.6\times10^{-19})^2}{5\times10^6\times1.6\times10^{-19}}=5.3\times10^{-14}\text{ m}\approx5\times10^{-12}\text{ cm} \).

Teacher's Note:
1. At the closest approach all the kinetic energy has become electric potential energy.
2. The result is far larger than the nuclear radius (a few fermi).

Question 23: An electron in an excited state of a hydrogen-like atom has orbit radius \( 4.5a_0 \) and orbital angular momentum \( \dfrac{3h}{2\pi} \). Which of the following could be wavelengths emitted when the atom de-excites (\( R \) = Rydberg constant)? (1) \( \dfrac{9}{32R} \) (2) \( \dfrac{9}{16R} \) (3) \( \dfrac{9}{5R} \) (4) \( \dfrac{4}{3R} \).
(a) (2) and (4) only
(b) (1) and (3) only
(c) (1) and (4) only
(d) (2) and (3) only
Show Answer & Explanation

Answer: (b) (1) and (3) only

Explanation:
1. Angular momentum \( n\dfrac{h}{2\pi} \) gives \( n=3 \). \( r=\dfrac{n^2a_0}{Z}=4.5a_0 \) gives \( Z=2 \) (He\( ^+ \)).
2. From \( n=3 \): \( 3\to1:\ \dfrac1\lambda=4R\left(1-\dfrac19\right)=\dfrac{32R}{9} \), so \( \lambda=\dfrac{9}{32R} \).
3. \( 3\to2:\ \dfrac1\lambda=4R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{9} \), so \( \lambda=\dfrac9{5R} \); and \( 2\to1:\ \lambda=\dfrac{1}{3R} \). So (1) and (3) occur.

Teacher's Note:
1. Find n from the angular momentum and Z from the radius.
2. List all possible transitions from the level.

Question 24: An electron beam is accelerated and then used to excite hydrogen. A neutron beam in which every neutron has the same kinetic energy is passed through a sample of a hydrogen-like gas (not hydrogen) in the ground state, producing six spectral lines, one of wavelength \( \dfrac{6200}{51}\text{ nm} \). The gas, and the minimum kinetic energy of the neutrons for this to happen, are:
(a) He\( ^+ \) and 63.75 eV
(b) Li\( ^{2+} \) and 63.75 eV
(c) He\( ^+ \) and 51 eV
(d) D and 54.4 eV
Show Answer & Explanation

Answer: (a) He\( ^+ \) and 63.75 eV

Explanation:
1. Six lines mean the ions reach \( n=4 \). The line \( \lambda=121.6\text{ nm} \) has energy \( 10.2\text{ eV} \); for \( Z=2 \): \( 54.4\left(\dfrac14-\dfrac1{16}\right)=10.2\text{ eV} \) (4 to 2), so the gas is He\( ^+ \).
2. Excitation energy \( 54.4\left(1-\dfrac1{16}\right)=51\text{ eV} \). In an inelastic collision with the ion (mass 4 times the neutron mass) only the fraction \( \dfrac{M}{m+M}=\dfrac45 \) of the neutron energy is available: \( K=\dfrac{51}{0.8}=63.75\text{ eV} \).

Teacher's Note:
1. Only energy in the centre-of-mass frame can excite the ion.
2. Six lines correspond to four levels.

Question 25: The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum is nearly:
(a) 802 nm
(b) 1882 nm
(c) 1648 nm
(d) 823 nm
Show Answer & Explanation

Answer: (d) 823 nm

Explanation:
1. The 122 nm line is the first Lyman line: \( \dfrac1{122}=\dfrac{3R}{4} \), so \( R=\dfrac{1}{91.5}\text{ nm}^{-1} \).
2. The infrared region begins with the Paschen series, whose smallest wavelength is its limit: \( \dfrac1\lambda=\dfrac R9 \), so \( \lambda=9\times91.5=823\text{ nm} \).

Teacher's Note:
1. The Paschen series (n = 3) is the first in the infrared.
2. The smallest wavelength of a series is its series limit.

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