JEE Physics Dual Nature Of Radiation And Matter MCQs Set 05

Physics Objective Questions and Answers: Dual Nature Of Radiation and Matter

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Multiple Choice Questions

Question 1: If E and P are the energy and the momentum of a photon, then on reducing the wavelength of the photon
(a) P and E both decrease
(b) P and E both increase
(c) P will increase and E decrease
(d) P will decrease and E increase
Show Answer & Explanation

Answer: (b) P and E both increase

Explanation:
1. \( E=hc/\lambda \) and \( P=h/\lambda \) both increase.

Teacher's Note:
1. Both vary inversely with wavelength.

Question 2: If the kinetic energy of a moving particle is E, the de Broglie wavelength is
(a) \( h\sqrt{2mE} \)
(b) \( \sqrt{2mE/h} \)
(c) \( h/\sqrt{2mE} \)
(d) \( hE/\sqrt{2mE} \)
Show Answer & Explanation

Answer: (c) \( h/\sqrt{2mE} \)

Explanation:
1. \( p=\sqrt{2mE} \) and \( \lambda=h/p \).

Teacher's Note:
1. lambda = h / p.

Question 3: An electron has energy 100 eV. Its wavelength will be
(a) 1.2 Å
(b) 10 Å
(c) 100 Å
(d) 1 Å
Show Answer & Explanation

Answer: (a) 1.2 Å

Explanation:
1. \( \lambda=12.27/\sqrt{100}=1.227 \) Å.

Teacher's Note:
1. lambda = 12.27 / sqrt(V) Å.

Question 4: The ratio of the wavelength of a deuteron and a proton accelerated through the same potential difference is
(a) \( 1/\sqrt2 \)
(b) \( \sqrt2/1 \)
(c) \( 1/2 \)
(d) \( 2/1 \)
Show Answer & Explanation

Answer: (a) \( 1/\sqrt2 \)

Explanation:
1. \( \lambda=h/\sqrt{2mqV} \), so \( \lambda\propto1/\sqrt m \).
2. The deuteron is twice as heavy.

Teacher's Note:
1. Same charge, double mass.

Question 5: An electron is accelerated from rest between two points A and B at potentials 20 V and 40 V. The de Broglie wavelength associated with the electron at B is
(a) 0.75 Å
(b) 7.5 Å
(c) 2.75 Å
(d) 2.75 m
Show Answer & Explanation

Answer: (c) 2.75 Å

Explanation:
1. The accelerating potential difference is 20 V.
2. \( \lambda=12.27/\sqrt{20}=2.74 \) Å.

Teacher's Note:
1. Use the potential difference.

Question 6: An electron moves with velocity \( 6.6\times10^{3} \) m/s. The de Broglie wavelength is (m = 9 x 10^-31 kg, h = 6.62 x 10^-34 J s)
(a) \( 1\times10^{-19}\ \text{m} \)
(b) \( 1\times10^{-5}\ \text{m} \)
(c) \( 1\times10^{-7}\ \text{m} \)
(d) \( 1\times10^{-10}\ \text{m} \)
Show Answer & Explanation

Answer: (c) \( 1\times10^{-7}\ \text{m} \)

Explanation:
1. \( \lambda=h/(mv)=6.62\times10^{-34}/(9\times10^{-31}\times6.6\times10^{3}) \).
2. \( \lambda\approx1.1\times10^{-7} \) m.

Teacher's Note:
1. lambda = h / m v.

Question 7: The energy to be added to an electron to reduce its de Broglie wavelength from \( 10^{-10} \) m to \( 0.5\times10^{-10} \) m is
(a) Four times the initial energy
(b) Equal to initial energy
(c) Twice the initial energy
(d) Thrice the initial energy
Show Answer & Explanation

Answer: (d) Thrice the initial energy

Explanation:
1. \( K\propto1/\lambda^2 \), so the final energy is 4 times the initial.
2. The added energy is 3 times the initial energy.

Teacher's Note:
1. Added energy = final minus initial.

Question 8: Electron, proton, neutron and alpha particle have the same kinetic energy of 1 MeV. In increasing order of wavelength
(a) \( \lambda_e,\lambda_p,\lambda_n,\lambda_\alpha \)
(b) \( \lambda_e,\lambda_n,\lambda_p,\lambda_\alpha \)
(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. \( \lambda=h/\sqrt{2mK} \), so the heaviest has the smallest wavelength.
2. Alpha is heavier than neutron, which is a little heavier than proton, then electron.

Teacher's Note:
1. Increasing wavelength means decreasing mass.

Question 9: The accelerating voltage of an electron gun is 50,000 V. The de Broglie wavelength of the electron is
(a) 0.55 Å
(b) 0.055 Å
(c) 0.077 Å
(d) 0.095 Å
Show Answer & Explanation

Answer: (b) 0.055 Å

Explanation:
1. \( \lambda=12.27/\sqrt{50000}=0.055 \) Å.

Teacher's Note:
1. Non-relativistic approximation.

Question 10: If the mass of a neutron is \( 1.7\times10^{-27} \) kg, the de Broglie wavelength of a neutron of energy 3 eV is
(a) \( 1.6\times10^{-10}\ \text{m} \)
(b) \( 1.6\times10^{-11}\ \text{m} \)
(c) \( 1.4\times10^{-10}\ \text{m} \)
(d) \( 1.4\times10^{-11}\ \text{m} \)
Show Answer & Explanation

Answer: (b) \( 1.6\times10^{-11}\ \text{m} \)

Explanation:
1. \( \lambda=h/\sqrt{2mK}=6.63\times10^{-34}/\sqrt{2\times1.7\times10^{-27}\times4.8\times10^{-19}} \).
2. \( \lambda=1.6\times10^{-11} \) m.

Teacher's Note:
1. Convert eV to joules.

Question 11: A proton and an alpha particle are accelerated through the same voltage. The ratio of their de Broglie wavelengths is
(a) 1 : 2
(b) \( 2\sqrt2:1 \)
(c) \( \sqrt2:1 \)
(d) 2 : 1
Show Answer & Explanation

Answer: (b) \( 2\sqrt2:1 \)

Explanation:
1. \( \lambda\propto1/\sqrt{mq} \). For alpha \( mq=4m_p\times2e \), eight times that of the proton.
2. So \( \lambda_p:\lambda_\alpha=\sqrt8:1=2\sqrt2:1 \).

Teacher's Note:
1. Alpha has mass 4m and charge 2e.

Question 12: The de Broglie wavelength of an atom at absolute temperature T is
(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 KE is \( \tfrac32kT \), so \( p=\sqrt{3mkT} \) and \( \lambda=h/\sqrt{3mkT} \).

Teacher's Note:
1. Thermal energy 3/2 kT.

Question 13: The de Broglie wavelength associated with the electron revolving in the ground state of a hydrogen atom is
(a) 0.3 Å
(b) 3.3 Å
(c) 6.62 Å
(d) 10 Å
Show Answer & Explanation

Answer: (b) 3.3 Å

Explanation:
1. \( \lambda=2\pi r_1=2\pi\times0.529 \) Å.
2. \( \lambda=3.3 \) Å.

Teacher's Note:
1. Orbit circumference for n = 1.

Question 14: The wavelength of a very fast moving electron (v nearly c) is
(a) \( \lambda=h/m_0v \)
(b) \( \lambda=h/\sqrt{2mE} \)
(c) \( \lambda^2=h^2/2mE \)
(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. With relativistic mass \( m=m_0/\sqrt{1-v^2/c^2} \) the wavelength \( h/mv \) gives the expression.

Teacher's Note:
1. Use relativistic mass.

Question 15: Which experiment explains the wave nature of electrons?
(a) Michelson's experiment
(b) Davisson-Germer experiment
(c) Roentgen experiment
(d) Rutherford experiment
Show Answer & Explanation

Answer: (b) Davisson-Germer experiment

Explanation:
1. Electron diffraction from a nickel crystal was observed in the Davisson-Germer experiment.

Teacher's Note:
1. Electron diffraction.

Question 16: In the Davisson-Germer experiment, the filament emits
(a) Photons
(b) Protons
(c) X-rays
(d) Electrons
Show Answer & Explanation

Answer: (d) Electrons

Explanation:
1. A heated filament emits electrons by thermionic emission.

Teacher's Note:
1. The beam is an electron beam.

Question 17: In the Davisson-Germer experiment, the diffracted waves are
(a) Electrons
(b) X-rays
(c) Photons
(d) Protons
Show Answer & Explanation

Answer: (a) Electrons

Explanation:
1. The electron waves are diffracted from the nickel crystal.

Teacher's Note:
1. Matter waves.

Question 18: 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. \( K=p^2/2m \) with the same p, so the lighter electron has the greater energy.

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

Question 19: The de Broglie equation for an electron shows
(a) Particle nature
(b) Wave nature
(c) Dual nature
(d) None of these
Show Answer & Explanation

Answer: (c) Dual nature

Explanation:
1. \( \lambda=h/p \) links the wave property \( \lambda \) to the particle property p.

Teacher's Note:
1. Dual nature.

Question 20: What happens to the de Broglie wavelength if the velocity of the electron is increased?
(a) It will increase
(b) It will decrease
(c) It will remain the same
(d) It will become twice
Show Answer & Explanation

Answer: (b) It will decrease

Explanation:
1. \( \lambda\propto1/v \).

Teacher's Note:
1. Inverse relation.

Question 21: Which of the following is true for a photon?
(a) \( E=hc/\lambda \)
(b) \( E=\tfrac12mv^2 \)
(c) \( P=E/2V \)
(d) \( E=\tfrac12mc^2 \)
Show Answer & Explanation

Answer: (a) \( E=hc/\lambda \)

Explanation:
1. The photon energy is \( E=h\nu=hc/\lambda \).

Teacher's Note:
1. Photon energy formula.

Question 22: For a moving particle of kinetic energy E, the correct de Broglie wavelength is
(a) Not applicable for a particle
(b) \( h/\sqrt{2mE} \)
(c) \( E\sqrt{h/2m} \)
(d) \( h/2mE \)
Show Answer & Explanation

Answer: (b) \( h/\sqrt{2mE} \)

Explanation:
1. \( \lambda=h/\sqrt{2mE} \).

Teacher's Note:
1. Standard formula.

Question 23: The de Broglie wavelength of an electron in the first Bohr orbit is
(a) Equal to the circumference of the first orbit
(b) Equal to twice the circumference
(c) Equal to half the circumference
(d) Equal to one fourth the circumference
Show Answer & Explanation

Answer: (a) Equal to the circumference of the first orbit

Explanation:
1. Bohr's quantisation condition requires an integral number of wavelengths, \( n\lambda=2\pi r \), so for n = 1 it equals the circumference.

Teacher's Note:
1. Standing wave on the orbit.

Question 24: If particles are moving with the same velocity, the maximum de Broglie wavelength is for
(a) Proton
(b) Alpha particle
(c) Neutron
(d) Beta particle
Show Answer & Explanation

Answer: (d) Beta particle

Explanation:
1. \( \lambda=h/mv \), so the lightest particle (beta, an electron) has the longest wavelength.

Teacher's Note:
1. Lightest means longest.

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Dual Nature Of Radiation and Matter Objective Questions & Solutions for JEE Physics

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