JEE Physics Modern Physics MCQs Set 03

Practice MCQs for JEE Physics Modern Physics

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Multiple Choice Questions

Question 1: Ionization potential of hydrogen is 13.6 eV. If it is excited by a photon of energy 12.1 eV, then the number of lines in the emission spectrum will be
(a) 2
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. \( 13.6 - 12.1 = 1.5 \) eV, so the atom reaches \( n = 3 \).
2. Number of lines \( = 3\times2/2 = 3 \).

Teacher's Note:
1. Count transitions down from \( n = 3 \).

Question 2: Consider the 4th orbit of He⁺ (helium), using non-relativistic approach, the speed of electron in this orbit will be (given K = 9 × 10⁹, Z = 2 and h = 6.6 × 10⁻³⁴ J s)
(a) \( 1.09\times10^6 \) m/s
(b) \( 0.73\times10^6 \) m/s
(c) \( 3.0\times10^8 \) m/s
(d) \( 2.92\times10^6 \) m/s
Show Answer & Explanation

Answer: (a) \( 1.09\times10^6 \) m/s

Explanation:
1. \( v = 2\pi KZe^2/(nh) = 2.18\times10^6\times Z/n m/s \).
2. \( v = 2.18\times10^6\times2/4 = 1.09\times10^6 m/s \).

Teacher's Note:
1. Speed grows with \( Z \) and falls with \( n \).

Question 3: Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line of the Lyman series in hydrogen spectrum will be
(a) \( 0.025\times10^4 \) m⁻¹
(b) \( 0.5\times10^7 \) m⁻¹
(c) \( 10^7 \) m⁻¹
(d) \( 2.5\times10^7 \) m⁻¹
Show Answer & Explanation

Answer: (c) \( 10^7 \) m⁻¹

Explanation:
1. The last line corresponds to \( n\to\infty \): \( \bar\nu = R(1 - 0) = R \).
2. The wave number is \( 10^7 m^{-1} \).

Teacher's Note:
1. The series limit gives the largest wave number.

Question 4: The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is
(a) 1 : 1
(b) 1 : −1
(c) 2 : −1
(d) 1 : −2
Show Answer & Explanation

Answer: (b) 1 : −1

Explanation:
1. \( K = -E \) for an electron in a Bohr orbit.
2. The ratio is 1 : −1.

Teacher's Note:
1. Total energy is negative of the kinetic energy.

Question 5: Characteristic X-rays are not obtained in the spectrum of H-atom because
(a) hydrogen is a gas
(b) hydrogen is very light
(c) energy difference in energy levels of hydrogen is much less
(d) energy difference in energy levels of hydrogen is much high
Show Answer & Explanation

Answer: (c) energy difference in energy levels of hydrogen is much less

Explanation:
1. Characteristic X-rays need transitions between inner shells with energy differences of keV.
2. In hydrogen the energy differences are only a few eV.

Teacher's Note:
1. Characteristic X-rays come from inner-shell vacancies of heavy atoms.

Question 6: For harder X-rays
(a) the wavelength is higher
(b) the intensity is higher
(c) the frequency is higher
(d) the photon energy is lower
Show Answer & Explanation

Answer: (c) the frequency is higher

Explanation:
1. Harder X-rays have higher photon energy, so they have higher frequency and a shorter wavelength.

Teacher's Note:
1. Hard means more penetrating.

Question 7: X-rays are produced in X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from
(a) 0 to ∞
(b) \( \lambda_{min} \) to ∞, where \( \lambda_{min}>0 \)
(c) 0 to \( \lambda_{max} \), where \( \lambda_{max}<\infty \)
(d) \( \lambda_{min} \) to \( \lambda_{max} \), where \( 0<\lambda_{min}<\lambda_{max}<\infty \)
Show Answer & Explanation

Answer: (b) \( \lambda_{min} \) to ∞, where \( \lambda_{min}>0 \)

Explanation:
1. The maximum photon energy is eV, so there is a minimum wavelength \( \lambda_{min} = hc/eV > 0 \).
2. There is no limit on the lower energy, so the wavelength extends to \( \infty \).

Teacher's Note:
1. The cut-off is only at the short wavelength end.

Question 8: The energy of a K-electron in tungsten is −20 keV and of an L-electron is −2 keV. The wavelength of X-rays emitted when there is electron jump from L to K shell is
(a) 0.3443 Å
(b) 0.6887 Å
(c) 1.3982 Å
(d) 4.78 Å
Show Answer & Explanation

Answer: (b) 0.6887 Å

Explanation:
1. \( \Delta E = 18 \) keV.
2. \( \lambda = 12.4/18 = 0.689 \) Å.

Teacher's Note:
1. Use λ(Å) = 12.4/E(keV).

Question 9: The wavelength of Kα line characteristic X-rays emitted by an element is 0.32 Å. The wavelength of Kβ line emitted by the same element will be
(a) 0.27 Å
(b) 0.32 Å
(c) 0.39 Å
(d) 0.2 Å
Show Answer & Explanation

Answer: (a) 0.27 Å

Explanation:
1. \( K\beta \) comes from the \( M \) to \( K \) transition, which has a larger energy than \( L \) to \( K \).
2. Its wavelength is shorter than 0.32 Å, about 0.27 Å (the ratio is \( (1-1/4)/(1-1/9)\approx0.84 \)).

Teacher's Note:
1. A bigger jump gives a shorter wavelength.

Question 10: The ionisation energies of K-shell for cobalt, copper, and molybdenum are 7.8, 9.0 and 20.1 keV respectively. If any metal out of these is used as target in an X-ray tube operated at 15 KV, then
(a) the K-series of characteristic X-ray will be emitted by Co only
(b) the K-series of characteristic X-rays will be emitted by Cu and Co only
(c) the K-series of characteristic X-rays will be emitted by Cu, Co and Mo
(d) the minimum wavelength of continuous X-rays emitted by the three metals will not be the same
Show Answer & Explanation

Answer: (b) the K-series of characteristic X-rays will be emitted by Cu and Co only

Explanation:
1. A K-shell vacancy needs electron energy above the \( K \) ionisation energy.
2. The 15 keV electrons exceed 7.8 and 9.0 keV, but not 20.1 keV.
3. So only Co and Cu emit the K-series.

Teacher's Note:
1. Compare the beam energy with the ionisation energy.

Question 11: What is the approximate ratio of wavelength of radiation Kα₁ and Kα₂ when the atomic number of first target is Z₁ = 64 and of second target is Z₂ = 80?
(a) 1/4
(b) 1/16
(c) \( 2/\sqrt5 \)
(d) 25/16
Show Answer & Explanation

Answer: (d) 25/16

Explanation:
1. Moseley's law: \( \lambda\propto1/(Z-1)^2 \).
2. \( \lambda_1/\lambda_2 = (79/63)^2\approx1.57\approx25/16 \).

Teacher's Note:
1. Use \( Z - 1 \) rather than \( Z \).

Question 12: If the nucleus Os¹⁸⁹ has a nuclear radius of about R, then which would have its radius approximately R/3?
(a) Li⁷
(b) Na²¹
(c) Fe⁵⁶
(d) Na²³
Show Answer & Explanation

Answer: (a) Li⁷

Explanation:
1. \( R\propto A^{1/3} \), so \( A = 189/27 = 7 \).
2. The nucleus is Li⁷.

Teacher's Note:
1. Radius scales as the cube root of \( A \).

Question 13: If the nuclear radius of ²⁷Al is 3.6 Fermi, the approximate nuclear radius of ⁶⁴Cu in Fermi is
(a) 4.8
(b) 3.6
(c) 2.4
(d) 1.2
Show Answer & Explanation

Answer: (a) 4.8

Explanation:
1. \( R = 3.6(64/27)^{1/3} = 3.6\times4/3 \).
2. \( R = 4.8 \) fermi.

Teacher's Note:
1. Use \( R ∝ A^{1/3} \).

Question 14: Let F_pp, F_pn and F_nn denote the nuclear force between proton–proton, proton–neutron and neutron–neutron pair respectively. When separation is 1 fm then
(a) F_pp < F_pn = F_nn
(b) F_pp > F_pn = F_nn
(c) F_pp = F_pn = F_nn
(d) F_pp < F_pn < F_nn
Show Answer & Explanation

Answer: (c) F_pp = F_pn = F_nn

Explanation:
1. Nuclear force is charge independent.
2. The three forces are equal at the same separation.

Teacher's Note:
1. Nuclear force does not distinguish protons from neutrons.

Question 15: The binding energy per nucleon for the parent nucleus is E₁ and that for the daughter nuclei is E₂. Then
(a) E₁ = 2E₂
(b) E₂ = 2E₁
(c) E₁ > E₂
(d) E₂ > E₁
Show Answer & Explanation

Answer: (d) E₂ > E₁

Explanation:
1. In fission energy is released, so the daughter nuclei are more tightly bound.
2. Hence \( E_2 > E_1 \).

Teacher's Note:
1. Higher binding energy per nucleon means more stable.

Question 16: When a nucleus with atomic number Z and mass number A undergoes a radioactive decay process, which of the following statement is incorrect?
(a) Both Z and A will decrease, if the process is α decay
(b) Z will decrease but A will not change, if the process is β⁺ decay
(c) Z will decrease but A will not change, if the process is β⁻ decay
(d) Z and A will remain unchanged, if the process is γ decay
Show Answer & Explanation

Answer: (c) Z will decrease but A will not change, if the process is β⁻ decay

Explanation:
1. In β⁻ decay a neutron becomes a proton, so \( Z \) increases by 1, not decreases.
2. The other three statements are right.

Teacher's Note:
1. β⁻ raises \( Z, \beta^{+} \) lowers \( Z \).

Question 17: Let mₚ be the mass of proton, mₙ the mass of neutron. M₁ the mass of ²⁰₁₀Ne nucleus and M₂ the mass of ⁴⁰₂₀Ca. Then
(a) M₂ = 2M₁
(b) M₂ > 2M₁
(c) M₂ < 2M₁
(d) M₁ < 10(mₙ + mₚ)
Show Answer & Explanation

Answer: (c) M₂ < 2M₁

Explanation:
1. Ca-40 has a higher binding energy per nucleon than Ne-20, so its mass defect is more than twice that of Ne-20.
2. Therefore \( M_2 < 2M_1 \).

Teacher's Note:
1. Mass defect grows faster than the number of nucleons.

Question 18: In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is: (A) E(²³⁶₉₂U) > E(¹³⁷₅₃I) + E(⁹⁷₃₉Y) + 2E(n); (B) E(²³⁶₉₂U) < E(¹³⁷₅₃I) + E(⁹⁷₃₉Y) + 2E(n); (C) E(²³⁶₉₂U) < E(¹⁴⁰₅₆Ba) + E(⁹⁴₃₆Kr) + 2E(n); (D) E(²³⁶₉₂U) > E(¹⁴⁰₅₆Ba) + E(⁹⁴₃₆Kr) + 2E(n)
(a) A & B
(b) A & D
(c) B & C
(d) C & D
Show Answer & Explanation

Answer: (b) A & D

Explanation:
1. Fission releases energy, so the rest mass energy of \( U \) is greater than that of the products.
2. \( (A) \) and \( (D) \) state this.

Teacher's Note:
1. Energy released equals the drop in rest mass energy.

Question 19: Which of the following is incorrect for a nuclear reaction?
(a) A typical fission may be represented by \( ^{235}_{92}U + ^1_0n \to ^{143}_{56}Ba + ^{93}_{36}Kr + \text{energy} \)
(b) Heavy water is used as moderator in preference to ordinary water because H may capture neutrons, while D does not
(c) Cadmium rods increase the reactor power when they go in, decrease when they go outward
(d) Slower neutrons are more effective in causing fission than faster neutrons in case of U²³⁵
Show Answer & Explanation

Answer: (c) Cadmium rods increase the reactor power when they go in, decrease when they go outward

Explanation:
1. Cadmium rods absorb neutrons.
2. Pushing them in decreases the chain reaction and the power, and pulling them out increases it, so statement 3 is reversed.

Teacher's Note:
1. Control rods absorb neutrons.

Question 20: The mass of a ⁴₂He nucleus is 0.0305 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of ⁴₂He nucleus is nearly
(a) 23 MeV
(b) 46 MeV
(c) 7.1 MeV
(d) 28.4 MeV
Show Answer & Explanation

Answer: (c) 7.1 MeV

Explanation:
1. \( B = 0.0305\times931.5 = 28.4 \) MeV.
2. Per nucleon \( = 28.4/4 = 7.1 \) MeV.

Teacher's Note:
1. Divide the total binding energy by \( A \).

Question 21: Mₚ denotes the mass of a proton and Mₙ that of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is the velocity of light)
(a) \( M(N,Z) = NM_n + ZM_p + Bc^2 \)
(b) \( M(N,Z) = NM_n + ZM_p + B/c^2 \)
(c) \( M(N,Z) = NM_n + ZM_p - B/c^2 \)
(d) \( M(N,Z) = NM_n + ZM_p - Bc^2 \)
Show Answer & Explanation

Answer: (c) \( M(N,Z) = NM_n + ZM_p - B/c^2 \)

Explanation:
1. The nucleus is lighter than its separated nucleons by the mass equivalent of the binding energy.
2. \( M = NM_n + ZM_p - B/c^2 \).

Teacher's Note:
1. Binding energy is released, so mass decreases.

Question 22: In the reaction ²₁H + ³₁H → ⁴₂He + ¹₀n, if the binding energies of ²₁H, ³₁H and ⁴₂He are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is
(a) a + b + c
(b) c + a − b
(c) c − a − b
(d) a + b − c
Show Answer & Explanation

Answer: (c) c − a − b

Explanation:
1. \( Q = (\text{BE of products}) - (\text{BE of reactants}) \).
2. The neutron has no binding energy: \( Q = c - (a + b) \).

Teacher's Note:
1. Energy released is the gain in total binding energy.

Question 23: For the given reaction, the particle X is: \( ^{14}_6C \to ^{14}_7N + \beta^- + X \)
(a) neutron
(b) anti neutrino
(c) neutrino
(d) proton
Show Answer & Explanation

Answer: (b) anti neutrino

Explanation:
1. β⁻ decay always emits an antineutrino with the electron, to conserve lepton number and energy.

Teacher's Note:
1. β⁻ is accompanied by an antineutrino.

Question 24: A nucleus with mass number 220 initially at rest emits an α-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the α-particle.
(a) 4.4 MeV
(b) 5.4 MeV
(c) 5.6 MeV
(d) 6.5 MeV
Show Answer & Explanation

Answer: (b) 5.4 MeV

Explanation:
1. Momentum conservation shares \( Q \) inversely as the masses.
2. \( K_\alpha = Q\times\dfrac{A-4}{A} = 5.5\times\dfrac{216}{220} = 5.4 \) MeV.

Teacher's Note:
1. The lighter particle takes the larger share.

Question 25: The binding energies per nucleon of ²₁D, ⁴₂He and ⁷₃Li are 1 MeV, 7 MeV and 5.5 MeV respectively. The energy released in the fusion reaction ²₁D + ⁷₃Li → 2 ⁴₂He + ¹₀n is about
(a) 20 MeV
(b) 16 MeV
(c) 8 MeV
(d) 1.6 MeV
Show Answer & Explanation

Answer: (b) 16 MeV

Explanation:
1. Binding energy of reactants \( = 2(1) + 7(5.5) = 40.5 \) MeV.
2. Products \( = 2\times4\times7 = 56 \) MeV.
3. Energy released \( = 56 - 40.5\approx16 \) MeV.

Teacher's Note:
1. Subtract total binding energies.

Question 26: A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be
(a) \( \dfrac{A-Z-4}{Z-2} \)
(b) \( \dfrac{A-Z-8}{Z-4} \)
(c) \( \dfrac{A-Z-4}{Z-8} \)
(d) \( \dfrac{A-Z-12}{Z-4} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{A-Z-4}{Z-8} \)

Explanation:
1. Mass number becomes \( A - 12 \). Atomic number becomes \( Z - 6 - 2 = Z - 8 \).
2. Neutrons \( = (A-12)-(Z-8) = A-Z-4 \).
3. Ratio \( = (A-Z-4)/(Z-8) \).

Teacher's Note:
1. Each positron lowers \( Z \) by 1.

Question 27: A nuclear fission is given as \( A^{240}\to B^{100} + C^{140} + Q \). Let binding energy per nucleon of nucleus A, B and C are 7.6 MeV, 8.1 MeV and 8.1 MeV respectively. The value of Q is (approximately)
(a) 120 MeV
(b) 220 MeV
(c) 240 MeV
(d) 180 MeV
Show Answer & Explanation

Answer: (a) 120 MeV

Explanation:
1. \( Q = (100+140)(8.1) - 240(7.6) \).
2. \( = 1944 - 1824 = 120 \) MeV.

Teacher's Note:
1. Gain in total binding energy.

Question 28: For the nuclear reaction \( ^{235}_{92}U + ^1_0n \to ^{144}_{56}Ba + \ldots + 3\,^1_0n \) the missing nucleus is
(a) \( ^{89}_{26}Kr \)
(b) \( ^{89}_{36}Kr \)
(c) \( ^{90}_{26}Kr \)
(d) \( ^{89}_{36}Sr \)
Show Answer & Explanation

Answer: (b) \( ^{89}_{36}Kr \)

Explanation:
1. Mass number: \( 236 = 144 + A + 3 \), so \( A = 89 \).
2. Atomic number: \( 92 = 56 + Z \), so \( Z = 36 \), which is krypton.

Teacher's Note:
1. Balance \( A \) and \( Z \) separately.

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