JEE Physics Dual Nature Of Radiation And Matter MCQs Set 18

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

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Multiple Choice Questions

Question 1: The maximum velocity of an electron emitted by light of wavelength λ incident on the surface of a metal of work-function φ is
(a) \( \sqrt{\frac{2(hc+\lambda\phi)}{m\lambda}} \)
(b) \( \frac{2(hc+\lambda\phi)}{m\lambda} \)
(c) \( \sqrt{\frac{2(hc-\lambda\phi)}{m\lambda}} \)
(d) \( \frac{2(h\lambda-\phi)}{m} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{\frac{2(hc-\lambda\phi)}{m\lambda}} \)

Explanation:
1. ½mv² = hc/λ − φ.
2. v = √(2(hc − λφ)/(mλ)).

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

Question 2: The graph of V₀ (stopping potential) versus ν (frequency) is
(a) a straight line with slope φ₀/e
(b) a straight line with slope −h/ν₀
(c) a straight line with slope −h/e
(d) a straight line with slope h/e
Show Answer & Explanation

Answer: (d) a straight line with slope h/e

Explanation:
1. eV₀ = hν − φ₀, so V₀ = (h/e)ν − φ₀/e, slope h/e.

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

Question 3: The theory of quantisation of electric charge was given by
(a) William Crookes
(b) J. J. Thomson
(c) R.A. Millikan
(d) Wilhelm Hallwachs
Show Answer & Explanation

Answer: (c) R.A. Millikan

Explanation:
1. Millikan's oil-drop experiment showed that charge is quantised.

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

Question 4: In a photon-particle collision, the quantity that does not remain conserved is
(a) total energy
(b) total momentum
(c) number of photons
(d) None of these
Show Answer & Explanation

Answer: (c) number of photons

Explanation:
1. A photon may be absorbed or created, so photon number is not conserved.

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

Question 5: The momentum of a photon of wavelength λ is
(a) hλ
(b) h/λ
(c) λ/h
(d) h/cλ
Show Answer & Explanation

Answer: (b) h/λ

Explanation:
1. p = h/λ.

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

Question 6: A 200 W street lamp emits yellow light of wavelength 0.6 μm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
(a) 1.5 × 10²⁰
(b) 6 × 10¹⁸
(c) 62 × 10²⁰
(d) 3 × 10¹⁹
Show Answer & Explanation

Answer: (a) 1.5 × 10²⁰

Explanation:
1. Light power = 50 W.
2. Photon energy = hc/λ = 3.3 × 10⁻¹⁹ J.
3. N = 50/3.3 × 10⁻¹⁹ ≈ 1.5 × 10²⁰ per second.

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

Question 7: If a photon and an electron have the same de-Broglie wavelength, then
(a) both have same kinetic energy
(b) proton has more K.E. than electron
(c) electron has more K.E. than proton
(d) both have same velocity
Show Answer & Explanation

Answer: (c) electron has more K.E. than proton

Explanation:
1. Same λ means the same momentum p. For a particle K = p²/2m, so the lighter electron has more K.E. than a proton.

Teacher's Note:
1. Compare kinetic energy at equal momentum.

Question 8: If the momentum of an electron is changed by P, then the de Broglie wavelength associated with it changes by 0.5%. The initial momentum of the electron will be
(a) \( 200P \)
(b) \( 400P \)
(c) \( \frac{P}{200} \)
(d) \( 100P \)
Show Answer & Explanation

Answer: (a) \( 200P \)

Explanation:
1. λ = h/p, so Δλ/λ = Δp/p = 0.5% = 1/200.
2. p = 200P.

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

Question 9:
Assertion (A): Photosensitivity of a metal is high if its work function is small.
Reason (R): Work function = hν₀ where ν₀ is the threshold frequency.
(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. A small work function means a low threshold frequency, so even low-energy light emits electrons.
2. The relation φ = hν₀ explains this.

Teacher's Note:
1. Link work function to threshold frequency.

Question 10: Match the Columns I and II.
Column IColumn II
(A) Field emission(1) Heat is supplied to the metal surface
(B) Photoelectric emission(2) Electric field is applied to the metal surface
(C) Thermionic emission(3) Light of suitable frequency illuminates the metal surface
(D) Secondary emission(4) Striking fast moving electrons on the metal surface

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

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

Explanation:
1. Field emission needs a strong electric field.
2. Photoelectric emission needs light.
3. Thermionic emission needs heating.
4. Secondary emission is due to fast electrons striking the surface.

Teacher's Note:
1. Link each emission type with its cause.

Question 11: Which of the following is/are false regarding cathode rays? I. They produce heating effect II. They don't deflect in electric field III. They cast shadow IV. They produce fluorescence
(a) I only
(b) II only
(c) I, II and III
(d) I, II, III and IV
Show Answer & Explanation

Answer: (b) II only

Explanation:
1. Cathode rays are charged electrons, so they do deflect in an electric field; II is false.

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

Question 12: 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. v = √(2eV/m) increases with accelerating potential.

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

Question 13: Which of the following experiments confirms the wave nature of electron?
(a) Millikan's oil drop experiment
(b) Davisson and Germer experiment
(c) Young's double slit experiment
(d) Geiger-Marsden experiment
Show Answer & Explanation

Answer: (b) Davisson and Germer experiment

Explanation:
1. Davisson-Germer observed electron diffraction.

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

Question 14: Match the Columns I and II.
Column IColumn II
(A) Einstein photoelectric equation(1) λ = h/p
(B) de-Broglie relation(2) Kmax = hν − φ₀
(C) Threshold frequency(3) Δx Δp ≈ h
(D) Heisenberg's uncertainty principle(4) ν = φ₀/h

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

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

Explanation:
1. Einstein's equation is Kmax = hν − φ₀.
2. de Broglie: λ = h/p.
3. Threshold frequency is ν₀ = φ₀/h.
4. Heisenberg: ΔxΔp ≈ h.

Teacher's Note:
1. Recall the standard formula of each name.

Question 15: The work functions of silver and sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is
(a) 1
(b) 2
(c) 4
(d) zero
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. V₀ = (h/e)ν − φ/e, so the slope h/e is the same for all metals.
2. Ratio = 1.

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

Question 16: In photoelectric effect, electrons are ejected from metals, if the incident light has a certain minimum
(a) wavelength
(b) frequency
(c) amplitude
(d) angle of incidence
Show Answer & Explanation

Answer: (b) frequency

Explanation:
1. The minimum frequency is the threshold frequency.

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

Question 17: A photon of 1.7 × 10⁻¹³ joule is absorbed by a material under special circumstances. The correct statement is
(a) Electrons of the atom of absorbed material will go the higher energy states
(b) Electron and positron pair will be created
(c) Only positron will be produced
(d) Photoelectric effect will occur and electron will be produced
Show Answer & Explanation

Answer: (b) Electron and positron pair will be created

Explanation:
1. 1.7 × 10⁻¹³ J ≈ 1.06 MeV, which exceeds 2mc² = 1.02 MeV.
2. So electron-positron pair production is possible.

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

Question 18: According to photoelectric equation Kmax = hν − φ₀, the photoelectric emission will not be possible if I. Kmax is negative II. ν₀ > ν III. Kmax is positive IV. ν₀ < ν
(a) I and II
(b) I and IV
(c) II and III
(d) III and IV
Show Answer & Explanation

Answer: (a) I and II

Explanation:
1. No emission when hν < φ₀, that is Kmax negative or ν below the threshold ν₀.

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

Question 19: In which of the following, emission of electrons does not take place?
(a) Thermionic emission
(b) X-rays emission
(c) Photoelectric emission
(d) Secondary emission
Show Answer & Explanation

Answer: (b) X-rays emission

Explanation:
1. X-rays are produced when fast electrons hit a target; they are photons, not electrons.

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

Question 20: The stopping potential is directly related to
(a) the work function of the metal
(b) intensity of incident radiation
(c) the saturation current for the given frequency
(d) the kinetic energy gained by the photoelectrons
Show Answer & Explanation

Answer: (d) the kinetic energy gained by the photoelectrons

Explanation:
1. eV₀ = Kmax, the maximum kinetic energy of the photoelectrons.

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

Question 21:
Assertion (A): The kinetic energy of photoelectrons emitted from a metal surface does not depend on the intensity of incident photon.
Reason (R): The ejection of electrons from the metallic surface is not possible with frequency of incident photons below the threshold frequency.
(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: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Kmax = hν − φ does not involve intensity, so the assertion is true.
2. The threshold fact is true but does not explain it.

Teacher's Note:
1. Both true; the reason is a different fact.

Question 22: 4 eV is the energy of the incident photon and the work function is 2 eV. The stopping potential will be
(a) \( 2\ \text{V} \)
(b) \( 4\ \text{V} \)
(c) \( 6\ \text{V} \)
(d) \( 2\sqrt2\ \text{V} \)
Show Answer & Explanation

Answer: (a) \( 2\ \text{V} \)

Explanation:
1. Kmax = 4 − 2 = 2 eV, so V₀ = 2 V.

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

Question 23: In Einstein's picture of photoelectric emission, the photoelectric emission does not take place by
(a) continuous emission of energy from radiation
(b) continuous absorption of energy from radiation
(c) discrete absorption of energy from radiation
(d) discrete emission of energy from radiation
Show Answer & Explanation

Answer: (b) continuous absorption of energy from radiation

Explanation:
1. Energy is absorbed in discrete quanta, not continuously.

Teacher's Note:
1. Einstein's picture is about discrete absorption of quanta.

Question 24: A steel ball of mass m is moving with a kinetic energy K. The de-Broglie wavelength associated with the ball is
(a) \( \frac{h}{2mK} \)
(b) \( \sqrt{\frac{h}{2mK}} \)
(c) \( \frac{h}{\sqrt{2mK}} \)
(d) \( \text{None of these} \)
Show Answer & Explanation

Answer: (c) \( \frac{h}{\sqrt{2mK}} \)

Explanation:
1. p = √(2mK), so λ = h/√(2mK).

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

Question 25: In Davisson-Germer experiment, an electron beam is incident on a crystal. The reflected beam consists of
(a) photons
(b) protons
(c) x-rays
(d) electrons
Show Answer & Explanation

Answer: (d) electrons

Explanation:
1. The scattered beam consists of electrons.

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

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