JEE Physics Dual Nature Of Radiation And Matter MCQs Set 19

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Chapter-wise Objective Questions: Dual Nature Of Radiation and Matter

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Multiple Choice Questions

Question 1: In Einstein's photoelectric equation Ek = hν − φ, Ek refers to
(a) kinetic energy of all the emitted electrons
(b) mean kinetic energy of emitted electrons
(c) maximum kinetic energy of emitted electrons
(d) minimum kinetic energy of emitted electrons
Show Answer & Explanation

Answer: (c) maximum kinetic energy of emitted electrons

Explanation:
1. Ek is the maximum kinetic energy.

Teacher's Note:
1. Recall the basic fact.

Question 2: If the energy of a photon is 10 eV, then its momentum is
(a) 5.33 × 10⁻²³ kg m/s
(b) 5.33 × 10⁻²⁵ kg m/s
(c) 5.33 × 10⁻²⁹ kg m/s
(d) 5.33 × 10⁻²⁷ kg m/s
Show Answer & Explanation

Answer: (d) 5.33 × 10⁻²⁷ kg m/s

Explanation:
1. p = E/c = 10 × 1.6 × 10⁻¹⁹/3 × 10⁸ = 5.33 × 10⁻²⁷ kg m/s.

Teacher's Note:
1. Recall the basic fact.

Question 3: It is essential to consider light as a stream of photons to explain
(a) diffraction of light
(b) refraction of light
(c) photoelectric effect
(d) reflection of light
Show Answer & Explanation

Answer: (c) photoelectric effect

Explanation:
1. Only the photoelectric effect cannot be explained by waves.

Teacher's Note:
1. Recall the basic fact.

Question 4: The photoelectric effect can be understood on the basis of
(a) wave theory of light only
(b) electromagnetic theory of light only
(c) quantum theory of light only
(d) None of these
Show Answer & Explanation

Answer: (c) quantum theory of light only

Explanation:
1. Einstein explained it with the quantum theory of light.

Teacher's Note:
1. Recall the basic fact.

Question 5: Radiations of two photon energies, twice and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of photoelectrons emitted in the two cases is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 1 : 1
Show Answer & Explanation

Answer: (b) 1 : 3

Explanation:
1. Kmax = φ and 9φ.
2. v ∝ √K, so the ratio is 1 : 3.

Teacher's Note:
1. Recall the basic fact.

Question 6: A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 2 m away, the number of electrons emitted by photocathode is reduced by a factor of
(a) 8
(b) 16
(c) 2
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. Intensity falls as 1/r².
2. At twice the distance it becomes 1/4, so the number of photoelectrons reduces by a factor of 4.

Teacher's Note:
1. Recall the basic fact.

Question 7: For photoelectric emission from a certain metal the cut-off frequency is ν. If radiation of frequency 2ν impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)
(a) \( \sqrt{h\nu/m} \)
(b) \( \sqrt{2h\nu/m} \)
(c) \( 2\sqrt{h\nu/m} \)
(d) \( \sqrt{h\nu/(2m)} \)
Show Answer & Explanation

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

Explanation:
1. Kmax = h(2ν) − hν = hν.
2. ½mv² = hν gives v = √(2hν/m).

Teacher's Note:
1. Recall the basic fact.

Question 8: The wavelength λe of an electron and λp of a photon of the same energy E are related by
(a) \( \lambda_p\propto\lambda_e \)
(b) \( \lambda_p\propto\sqrt{\lambda_e} \)
(c) \( \lambda_p\propto\frac{1}{\sqrt{\lambda_e}} \)
(d) \( \lambda_p\propto\lambda_e^2 \)
Show Answer & Explanation

Answer: (d) \( \lambda_p\propto\lambda_e^2 \)

Explanation:
1. Electron: λe = h/√(2mE), so E = h²/(2mλe²).
2. Photon: λp = hc/E = 2mcλe²/h, so λp ∝ λe².

Teacher's Note:
1. Recall the basic fact.

Question 9:
Assertion (A): In the process of photoelectric emission, all emitted electrons do not have the same kinetic energy.
Reason (R): The energy of electrons emitted from inside the metal surface is lost in collision with the other atoms in the metal.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Electrons deeper in the metal lose energy to collisions on the way out.
2. Hence electrons have a spread of kinetic energies from zero to Kmax.

Teacher's Note:
1. Energy loss inside the metal gives a spread of energies.

Question 10:
Assertion (A): Two sources of equal intensity always emit an equal number of photons in any time interval.
Reason (R): Two sources of equal intensity may emit an equal number of photons in any time interval.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Photon emission is a random process, so the numbers can differ in short intervals; the assertion is false.
2. The reason is true.

Teacher's Note:
1. Photon emission is statistical.

Question 11: Light of a particular frequency ν is incident on a metal surface. When the intensity of incident radiation is increased, the photoelectric current
(a) decreases
(b) increases
(c) remains unchanged
(d) sometimes increases and sometimes decreases
Show Answer & Explanation

Answer: (b) increases

Explanation:
1. More photons per second eject more electrons, so the photocurrent increases.

Teacher's Note:
1. Recall the basic fact.

Question 12: The wave nature of light was established by (i) Maxwell's equations (ii) Fraunhofer's lines (iii) Hertz experiment (iv) Einstein's theory
(a) (i) and (ii) only
(b) (ii) and (iv) only
(c) (i) and (iii) only
(d) (iii) and (iv) only
Show Answer & Explanation

Answer: (c) (i) and (iii) only

Explanation:
1. Maxwell's theory and Hertz's detection of electromagnetic waves established the wave nature of light.

Teacher's Note:
1. Recall the basic fact.

Question 13: A photosensitive metal is not emitting photo-electrons when irradiated. To cross the threshold we need to increase
(a) intensity
(b) frequency
(c) wavelength
(d) None of these
Show Answer & Explanation

Answer: (b) frequency

Explanation:
1. Increase the frequency above the threshold frequency.

Teacher's Note:
1. Recall the basic fact.

Question 14: The photoelectric threshold of tungsten is 2300 Å. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 1800 Å is
(a) 0.15 eV
(b) 1.5 eV
(c) 15 eV
(d) 150 eV
Show Answer & Explanation

Answer: (b) 1.5 eV

Explanation:
1. Photon energy = 12375/1800 = 6.875 eV.
2. φ = 12375/2300 = 5.38 eV.
3. Kmax ≈ 1.5 eV.

Teacher's Note:
1. Recall the basic fact.

Question 15: If E₁, E₂, E₃ are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
(a) E₁ > E₃ > E₂
(b) E₂ > E₃ > E₁
(c) E₁ > E₂ > E₃
(d) E₁ = E₂ = E₃
Show Answer & Explanation

Answer: (a) E₁ > E₃ > E₂

Explanation:
1. Same λ means same momentum, and K = p²/2m.
2. Smaller mass gives larger K: electron > proton > alpha.

Teacher's Note:
1. Recall the basic fact.

Question 16: Match the quantities in Column I with their definitions in Column II.
Column IColumn II
(A) Photocurrent(1) The minimum energy required by an electron to escape from the metal surface
(B) Saturation current(2) The minimum retarding potential
(C) Stopping potential(3) The number of photoelectrons emitted per second
(D) Work function(4) The maximum number of photoelectrons emitted per second

(a) A-3, B-4, C-2, D-1
(b) A-3, B-4, C-1, D-2
(c) A-3, B-1, C-2, D-4
(d) A-2, B-4, C-3, D-1
Show Answer & Explanation

Answer: (a) A-3, B-4, C-2, D-1

Explanation:
1. Photocurrent corresponds to electrons emitted per second.
2. Saturation current is the maximum such rate.
3. Stopping potential is the minimum retarding potential.
4. Work function is the minimum escape energy.

Teacher's Note:
1. Match each quantity with its definition.

Question 17: A photoelectric cell is a device which
(a) converts light into electricity
(b) converts electricity into light
(c) stores light
(d) stores electricity
Show Answer & Explanation

Answer: (a) converts light into electricity

Explanation:
1. It produces current when illuminated.

Teacher's Note:
1. Recall the basic fact.

Question 18: The magnitude of the de-Broglie wavelength (λ) of electron (e), proton (p), neutron (n) and α-particle (α) all having the same energy of 1 MeV, in the increasing order will follow the sequence
(a) λe, λp, λn, λα
(b) λe, λn, λp, λα
(c) λα, λn, λp, λe
(d) λp, λe, λα, λn
Show Answer & Explanation

Answer: (c) λα, λn, λp, λe

Explanation:
1. λ = h/√(2mK) is smaller for heavier particles.
2. Mass order: α > n > p > e, so λα < λn < λp < λe.

Teacher's Note:
1. Recall the basic fact.

Question 19: The work function of aluminium is 4.2 eV. If two photons, each of energy 3.5 eV strike an electron of aluminium, then emission of electrons
(a) will be possible
(b) will not be possible
(c) Data is incomplete
(d) Depends upon the density of the surface
Show Answer & Explanation

Answer: (b) will not be possible

Explanation:
1. An electron absorbs one photon at a time (3.5 eV < 4.2 eV).
2. The energies cannot be added, so no emission.

Teacher's Note:
1. Recall the basic fact.

Question 20: The particle nature of light is not confirmed by
(a) photoelectric effect
(b) scattering of X-rays by electrons
(c) diffraction of electrons
(d) Compton effect
Show Answer & Explanation

Answer: (c) diffraction of electrons

Explanation:
1. Diffraction of electrons shows the wave nature of matter.

Teacher's Note:
1. Recall the basic fact.

Question 21: Which of the following statements are true? I. In the interaction with matter, radiation behaves as if it is made up of particles called photons. II. Each photon has energy E = hν and momentum p = hν/c. III. Photons are electrically neutral and are not deflected by electric and magnetic fields. IV. In a photon-particle collision, photon number is conserved.
(a) I and II
(b) I, II and III
(c) I, III and IV
(d) I, II and IV
Show Answer & Explanation

Answer: (b) I, II and III

Explanation:
1. Photon number is not conserved in collisions, so IV is false.
2. I, II and III are true.

Teacher's Note:
1. Recall the basic fact.

Question 22: A photon will have less energy, if its
(a) amplitude is higher
(b) frequency is higher
(c) wavelength is longer
(d) wavelength is shorter
Show Answer & Explanation

Answer: (c) wavelength is longer

Explanation:
1. E = hc/λ is smaller for longer wavelength.

Teacher's Note:
1. Recall the basic fact.

Question 23: The photoelectric effect is based on the law of conservation of
(a) momentum
(b) energy
(c) angular momentum
(d) mass
Show Answer & Explanation

Answer: (b) energy

Explanation:
1. hν = φ + Kmax is an energy balance.

Teacher's Note:
1. Recall the basic fact.

Question 24: Photons are deflected by
(a) electric field only
(b) magnetic field only
(c) electromagnetic field
(d) None of these
Show Answer & Explanation

Answer: (d) None of these

Explanation:
1. Photons are neutral, so no electric or magnetic field deflects them.

Teacher's Note:
1. Recall the basic fact.

Question 25: In a photoelectric experiment the stopping potential for the incident light of wavelength 4000 Å is 2 volt. If the wavelength is changed to 3000 Å, the stopping potential will be
(a) 2 V
(b) zero
(c) less than 2 V
(d) more than 2 V
Show Answer & Explanation

Answer: (d) more than 2 V

Explanation:
1. Shorter wavelength means higher photon energy and higher Kmax.
2. So the stopping potential increases (about 3.0 V).

Teacher's Note:
1. Recall the basic fact.

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