JEE Physics Modern Physics MCQs Set 02

Multiple Choice Questions (MCQs) for JEE Physics: Modern Physics

Access targeted multiple-choice questions for Modern Physics designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Modern Physics MCQs for JEE Physics

Access the complete set of multiple-choice questions for Modern Physics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: Potential difference required to produce electrons of wavelength 5 Å in an electron microscope will be
(a) 2.5 V
(b) 6 V
(c) 2.5 kV
(d) 5 kV
Show Answer & Explanation

Answer: (b) 6 V

Explanation:
1. \( \lambda = 12.27/\sqrt V \) Å, so \( V = (12.27/5)^2\approx6 \) V.

Teacher's Note:
1. Use the electron wavelength formula.

Question 2: If the momentum of an electron is changed by P, then the de Broglie wavelength associated with it changes by 0.2%. The initial momentum of electron will be
(a) P/200
(b) 100 P
(c) 500 P
(d) 400 P
Show Answer & Explanation

Answer: (c) 500 P

Explanation:
1. \( \lambda = h/p \), so \( \Delta\lambda/\lambda = \Delta p/p \).
2. \( p = P/0.002 = 500P \).

Teacher's Note:
1. Fractional changes of λ and p are equal in size.

Question 3: An α-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m². The de Broglie wavelength associated with the particle will be
(a) 10 Å
(b) 0.1 Å
(c) 1 Å
(d) 0.01 Å
Show Answer & Explanation

Answer: (d) 0.01 Å

Explanation:
1. \( p = qBr = 2(1.6\times10^{-19})(0.25)(0.83\times10^{-2}) = 6.6\times10^{-22} \) kg·m/s.
2. \( \lambda = h/p = 6.63\times10^{-34}/6.6\times10^{-22} = 10^{-12} \) m = 0.01 Å.

Teacher's Note:
1. Momentum of a charged particle in a magnetic field is qBr.

Question 4: Electrons used in an electron microscope are accelerated by a voltage of 10 kV. If the voltage is increased to 40 kV then the de-Broglie wavelength associated with the electrons would
(a) increase by 2 times
(b) decrease by 2 times
(c) decrease by 4 times
(d) increase by 4 times
Show Answer & Explanation

Answer: (b) decrease by 2 times

Explanation:
1. \( \lambda\propto1/\sqrt V \).
2. The voltage is 4 times larger, so λ decreases by a factor of 2.

Teacher's Note:
1. The square root of 4 is 2.

Question 5: The de Broglie equation for an electron shows its
(a) particle nature
(b) wave nature
(c) dual nature
(d) none of these
Show Answer & Explanation

Answer: (b) wave nature

Explanation:
1. The equation \( \lambda = h/p \) connects a wave property (λ) with a particle property (p).
2. It shows that the electron has a wave nature.

Teacher's Note:
1. De Broglie waves are matter waves.

Question 6: An electron and a proton have the same de Broglie wavelength. Then the kinetic energy of the electron is
(a) zero
(b) infinity
(c) equal to kinetic energy of proton
(d) greater than the kinetic energy of proton
Show Answer & Explanation

Answer: (d) greater than the kinetic energy of proton

Explanation:
1. Equal λ means equal momentum p.
2. \( K = p^2/2m \) is larger for the lighter electron.

Teacher's Note:
1. Same momentum, smaller mass, larger K.

Question 7: If an electron and a photon propagate with the same wavelength, it implies that they can have same
(a) energy
(b) momentum
(c) velocity
(d) angular momentum
Show Answer & Explanation

Answer: (b) momentum

Explanation:
1. Both have \( p = h/\lambda \), so equal wavelength means equal momentum.
2. Their energies and speeds differ.

Teacher's Note:
1. De Broglie relation applies to both.

Question 8: The ratio of momentum of an α-particle to the electron when both are accelerated from rest by a potential difference of 100 V, is
(a) 1
(b) \( \sqrt{2m_\alpha/m_e} \)
(c) \( \sqrt{m_e/m_\alpha} \)
(d) \( \sqrt{m_e/2m_\alpha} \)
Show Answer & Explanation

Answer: (b) \( \sqrt{2m_\alpha/m_e} \)

Explanation:
1. \( p = \sqrt{2mqV} \).
2. \( p_\alpha/p_e = \sqrt{2m_\alpha(2e)/(2m_ee)} = \sqrt{2m_\alpha/m_e} \).

Teacher's Note:
1. The α-particle has twice the charge.

Question 9: After absorbing a slowly moving neutron of mass mₙ (momentum ≈ 0) a nucleus of mass M breaks into two nuclei of masses m₁ and 5m₁ (6m₁ = M + mₙ), respectively. If the de Broglie wavelength of the nucleus with mass m₁ is λ, then de Broglie wavelength of the other nucleus will be
(a) 25λ
(b) 5λ
(c) λ/5
(d) λ
Show Answer & Explanation

Answer: (d) λ

Explanation:
1. The initial momentum is nearly zero, so the two fragments have equal and opposite momenta.
2. \( \lambda = h/p \) is the same for both: λ.

Teacher's Note:
1. Momentum is conserved.

Question 10: The magnitude of de Broglie wavelength (λ) of electron (e), proton (p), neutron (n) and α-particle (α) all having the same kinetic energy of 1 MeV, in the increasing order will follow the sequence
(a) \( \lambda_e<\lambda_p<\lambda_n<\lambda_\alpha \)
(b) \( \lambda_e<\lambda_\alpha<\lambda_n<\lambda_p \)
(c) \( \lambda_\alpha<\lambda_n<\lambda_p<\lambda_e \)
(d) \( \lambda_p<\lambda_e<\lambda_\alpha<\lambda_n \)
Show Answer & Explanation

Answer: (c) \( \lambda_\alpha<\lambda_n<\lambda_p<\lambda_e \)

Explanation:
1. \( \lambda = h/\sqrt{2mK} \) decreases as the mass increases.
2. The mass order is \( m_\alpha > m_n > m_p > m_e \).
3. So \( \lambda_\alpha < \lambda_n < \lambda_p < \lambda_e \).

Teacher's Note:
1. For equal K, the heavier particle has the smaller wavelength.

Question 11: The de Broglie wavelength of an atom at absolute temperature T K will be
(a) \( h/mkT \)
(b) \( h/\sqrt{3mkT} \)
(c) \( \sqrt{3mkT}/h \)
(d) \( \sqrt{3mkT} \)
Show Answer & Explanation

Answer: (b) \( h/\sqrt{3mkT} \)

Explanation:
1. Average kinetic energy \( = \tfrac32kT = p^2/2m \), so \( p = \sqrt{3mkT} \).
2. \( \lambda = h/\sqrt{3mkT} \).

Teacher's Note:
1. Use the thermal kinetic energy.

Question 12: The wavelength of very fast moving electron (v ≈ c) is
(a) \( \lambda = h/m_0v \)
(b) \( \lambda = h/\sqrt{2mE} \)
(c) \( \lambda = h^2/\sqrt{2mE} \)
(d) \( \lambda = \dfrac{h}{m_0v}\sqrt{1-v^2/c^2} \)
Show Answer & Explanation

Answer: (d) \( \lambda = \dfrac{h}{m_0v}\sqrt{1-v^2/c^2} \)

Explanation:
1. For speeds near c the momentum is \( p = m_0v/\sqrt{1-v^2/c^2} \).
2. \( \lambda = h/p = \dfrac h{m_0v}\sqrt{1-v^2/c^2} \).

Teacher's Note:
1. Use relativistic momentum.

Question 13: A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV, its kinetic energy is
(a) 1840 keV
(b) 1/1840 keV
(c) 1 keV
(d) 920 keV
Show Answer & Explanation

Answer: (c) 1 keV

Explanation:
1. \( K = qV = e\times1000 \) V = 1 keV.
2. The mass does not matter.

Teacher's Note:
1. Energy gained depends on charge and voltage only.

Question 14: What is the relation between the momentum p and the associated de Broglie wavelength λ of a particle?
(a) p is inversely proportional to λ, the graph of p against λ being a hyperbola
(b) p decreases linearly with λ
(c) p is directly proportional to λ
(d) p is independent of λ
Show Answer & Explanation

Answer: (a) p is inversely proportional to λ, the graph of p against λ being a hyperbola

Explanation:
1. \( \lambda = h/p \), so \( p = h/\lambda \).
2. The graph of p against λ is a rectangular hyperbola.

Teacher's Note:
1. Inverse proportion gives a hyperbola.

Question 15: According to de Broglie, wavelength of electron in second orbit is 10⁻⁹ metre. Then the circumference of orbit is
(a) 10⁻⁹ m
(b) \( 2\times10^{-9} \) m
(c) \( 3\times10^{-9} \) m
(d) \( 4\times10^{-9} \) m
Show Answer & Explanation

Answer: (b) \( 2\times10^{-9} \) m

Explanation:
1. Circumference \( = n\lambda = 2\times10^{-9} \) m.

Teacher's Note:
1. The orbit holds n whole wavelengths.

Question 16: In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by
(a) increasing the potential difference between the anode and filament
(b) increasing the filament current
(c) decreasing the filament current
(d) decreasing the potential difference between the anode and filament
Show Answer & Explanation

Answer: (a) increasing the potential difference between the anode and filament

Explanation:
1. \( \tfrac12mv^2 = eV \).
2. A larger anode-to-filament voltage gives a larger speed.

Teacher's Note:
1. The filament current changes only the number of electrons.

Question 17: In Davisson–Germer experiment, Nickel crystal acts as
(a) perfect reflector
(b) three dimensional diffraction grating
(c) ideal absorber
(d) two dimensional diffraction grating
Show Answer & Explanation

Answer: (b) three dimensional diffraction grating

Explanation:
1. The regular lattice of atoms diffracts the electron waves.
2. It acts as a three dimensional diffraction grating.

Teacher's Note:
1. The Bragg law is applied here.

Question 18: The velocity at which the mass of a particle becomes twice its rest mass will be
(a) \( 2c/3 \)
(b) \( c/2 \)
(c) \( c\sqrt3/2 \)
(d) \( 3c/4 \)
Show Answer & Explanation

Answer: (c) \( c\sqrt3/2 \)

Explanation:
1. \( m = m_0/\sqrt{1-v^2/c^2} = 2m_0 \).
2. \( 1 - v^2/c^2 = 1/4 \), so \( v = c\sqrt3/2 \).

Teacher's Note:
1. Use relativistic mass.

Question 19: Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be
(a) 3
(b) 2
(c) 6
(d) 10
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. Photon energy \( = 12400/975 = 12.72 \) eV, which excites the atom to n = 4 (12.75 eV).
2. Number of lines \( = 4\times3/2 = 6 \).

Teacher's Note:
1. Lines = n(n−1)/2.

Question 20: Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr's atomic model?
(a) 0.65 eV
(b) 1.9 eV
(c) 11.1 eV
(d) 13.6 eV
Show Answer & Explanation

Answer: (c) 11.1 eV

Explanation:
1. Possible values are differences of the levels \( 13.6(1/n_1^2 - 1/n_2^2) \): 0.66 eV (4→3), 1.89 eV (3→2), 13.6 eV (∞→1).
2. 11.1 eV is not any such difference.

Teacher's Note:
1. Check each against the allowed differences.

Question 21: The ground state energy of hydrogen atom is −13.6 eV. When its electron is in the first excited state, its excitation energy is
(a) 10.2 eV
(b) 0
(c) 3.4 eV
(d) 6.8 eV
Show Answer & Explanation

Answer: (a) 10.2 eV

Explanation:
1. The first excited state is n = 2 with energy −3.4 eV.
2. Excitation energy \( = -3.4 - (-13.6) = 10.2 \) eV.

Teacher's Note:
1. Excitation energy is measured from the ground state.

Question 22: According to the Bohr theory of Hydrogen atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number n, respectively, as
(a) \( \dfrac1n,\ \dfrac1{n^2},\ n^2 \)
(b) \( \dfrac1n,\ n^2,\ \dfrac1{n^2} \)
(c) \( n^2,\ \dfrac1{n^2},\ n^2 \)
(d) \( n,\ \dfrac1{n^2},\ \dfrac1{n^2} \)
Show Answer & Explanation

Answer: (a) \( \dfrac1n,\ \dfrac1{n^2},\ n^2 \)

Explanation:
1. \( v\propto1/n \), \( |E|\propto1/n^2 \), \( r\propto n^2 \).

Teacher's Note:
1. These three proportionalities are standard.

Question 23: In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
(a) \( 5/27 \)
(b) \( 4/9 \)
(c) \( 9/4 \)
(d) \( 27/5 \)
Show Answer & Explanation

Answer: (a) \( 5/27 \)

Explanation:
1. Lyman longest: \( 1/\lambda = R(1-1/4) = 3R/4 \).
2. Balmer longest: \( 1/\lambda = R(1/4-1/9) = 5R/36 \).
3. Ratio \( = (4/3)/(36/5) = 5/27 \).

Teacher's Note:
1. The longest wavelength is the first line of the series.

Question 24: Wavelength of photon released from transition of electron from 4 → 2 in He⁺ is equivalent to transition of electron in a hydrogen atom from
(a) 3 → 1
(b) 2 → 1
(c) 3 → 2
(d) 4 → 2
Show Answer & Explanation

Answer: (b) 2 → 1

Explanation:
1. For He⁺, \( \Delta E = 4\times13.6(1/4 - 1/16) = 10.2 \) eV.
2. For hydrogen the 2 → 1 transition gives \( 13.6(1 - 1/4) = 10.2 \) eV.

Teacher's Note:
1. Z² scaling relates the two atoms.

Question 25: An electron of a stationary hydrogen atom passes from the 4th energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be (m is the mass of the atom, R is the Rydberg constant and h is Planck's constant)
(a) \( \dfrac{15m}{16hR} \)
(b) \( \dfrac{24m}{25hR} \)
(c) \( \dfrac{15hR}{16m} \)
(d) \( \dfrac{25hR}{24m} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{15hR}{16m} \)

Explanation:
1. \( 1/\lambda = R(1-1/16) = 15R/16 \).
2. Photon momentum \( h/\lambda = 15hR/16 \).
3. Recoil speed \( = 15hR/(16m) \).

Teacher's Note:
1. Momentum conservation gives the recoil.

Question 26: An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be (m is the mass of the atom, R is the Rydberg constant and h is Planck's constant)
(a) \( \dfrac{25m}{24hR} \)
(b) \( \dfrac{24m}{25hR} \)
(c) \( \dfrac{24hR}{25m} \)
(d) \( \dfrac{25hR}{24m} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{24hR}{25m} \)

Explanation:
1. \( 1/\lambda = R(1-1/25) = 24R/25 \).
2. Recoil speed \( = h/(m\lambda) = 24hR/(25m) \).

Teacher's Note:
1. Use the same method as for the 4 → 1 jump.

Question 27: What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of Lyman series?
(a) 4 : 1
(b) 4 : 3
(c) 4 : 9
(d) 5 : 9
Show Answer & Explanation

Answer: (a) 4 : 1

Explanation:
1. Balmer shortest: \( 1/\lambda = R/4 \), so \( \lambda = 4/R \).
2. Lyman shortest: \( \lambda = 1/R \).
3. Ratio = 4 : 1.

Teacher's Note:
1. The shortest wavelength is the series limit.

Question 28: If the ionization potential of hydrogen atom is 13.6 eV, its energy in the n = 3 is approximately
(a) −1.14 eV
(b) −1.51 eV
(c) −3.4 eV
(d) −4.53 eV
Show Answer & Explanation

Answer: (b) −1.51 eV

Explanation:
1. \( E_3 = -13.6/9 = -1.51 \) eV.

Teacher's Note:
1. Energy goes as 1/n².

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