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Quiz Chapter 7: Quantum Physics

10 questions · Form 5 Physics Bab 7: Quantum Physics

Question 1 of 10Score: 0

According to Louis de Broglie, which physical property determines the de Broglie wavelength of a moving particle?

Full Question List & Answer Key

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1. According to Louis de Broglie, which physical property determines the de Broglie wavelength of a moving particle?

  1. Its electric charge
  2. Its momentum (p = mv)
  3. Its surface area
  4. Its temperature
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Answer: B

The de Broglie wavelength is given by λ = hp = hm × v.

2. What is the threshold frequency (f₀) of a metal?

  1. The maximum frequency of light that causes reflection
  2. The minimum frequency of light required to emit photoelectrons from a metal surface
  3. The frequency at which photoelectrons reach zero kinetic energy in a vacuum
  4. The frequency of light produced by thermal heating
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Answer: B

Threshold frequency (f₀) is the minimum photon frequency needed to overcome the metal's work function (Φ = hf₀).

3. In a photoelectric experiment, the stopping potential V_s is measured to be 1.5 V. What is the maximum kinetic energy of the photoelectrons in Joules? (e = 1.6 × 10⁻¹⁹ C)

  1. 2.4 × 10⁻¹⁹ J
  2. 1.07 × 10⁻¹⁹ J
  3. 1.50 J
  4. 3.84 × 10⁻¹⁹ J
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Answer: A

K_max = e × V_s = (1.6 × 10⁻¹⁹ C) × 1.5 V = 2.4 × 10⁻¹⁹ J.

4. What is a photon in modern quantum theory?

  1. A positively charged subatomic particle inside atomic nuclei
  2. A discrete packet of energy of electromagnetic radiation
  3. A high-speed electron emitted during beta decay
  4. An uncharged particle with infinite rest mass
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Answer: B

According to Max Planck and Albert Einstein, electromagnetic radiation travels in discrete energy packets called photons.

5. What happens to the stopping potential (V_s) if light of a higher frequency is used in a photoelectric setup?

  1. Stopping potential decreases
  2. Stopping potential increases
  3. Stopping potential remains unchanged
  4. Stopping potential drops to zero
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Answer: B

Higher photon frequency yields higher maximum kinetic energy (K_max), requiring a larger stopping potential (e V_s = K_max) to halt emitted photoelectrons.

6. A light source emits photons with a wavelength of 400 nm. What is the momentum of each photon? (h = 6.63 × 10⁻³⁴ J s)

  1. 1.66 × 10⁻²⁷ kg m s⁻¹
  2. 2.65 × 10⁻⁴⁰ kg m s⁻¹
  3. 1.66 × 10⁻²⁹ kg m s⁻¹
  4. 6.63 × 10⁻²⁷ kg m s⁻¹
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Answer: A

Momentum p = h / λ = 6.63 × 10⁻³⁴ J s400 × 10⁻⁹ m = 1.6575 × 10⁻²⁷ kg m s⁻¹ ≈ 1.66 × 10⁻²⁷ kg m s⁻¹.

7. Which graph correctly represents the relationship between the maximum kinetic energy (K_max) of photoelectrons and the frequency (f) of incident light?

  1. A straight line with a positive slope intersecting the horizontal frequency axis at threshold frequency f₀
  2. A horizontal straight line parallel to the frequency axis
  3. A curve starting from the origin increasing exponentially
  4. A straight line passing through the origin
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Answer: A

From K_max = hf - hf₀, the graph of K_max vs f is a straight line y = mx + c with gradient = h and x-intercept = f₀.

8. Why could classical wave theory NOT explain the photoelectric effect?

  1. Classical theory predicted that electron emission required continuous light wave energy buildup over time
  2. Classical theory predicted that light travels faster in dense media
  3. Classical theory predicted that electrons carry positive charges
  4. Classical theory stated that photons have zero momentum
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Answer: A

Wave theory assumed energy spread continuously over wave fronts, predicting time delays for dim light, whereas experiments proved emission is instantaneous.

9. The work function of Potassium is 3.68 × 10⁻¹⁹ J. Calculate its threshold frequency f₀. (h = 6.63 × 10⁻³⁴ J s)

  1. 5.55 × 10¹⁴ Hz
  2. 2.44 × 10⁻53 Hz
  3. 1.80 × 10¹⁵ Hz
  4. 5.55 × 10¹² Hz
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Answer: A

f₀ = Work Function / h = 3.68 × 10⁻¹⁹ J6.63 × 10⁻³⁴ J s = 5.55 × 10¹⁴ Hz.

10. Calculate the de Broglie wavelength of an electron moving at a velocity of 2.0 × 10⁶ m s⁻¹. (m_e = 9.1 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s)

  1. 3.64 × 10⁻¹⁰ m
  2. 1.82 × 10⁻²⁴ m
  3. 3.64 × 10⁻³⁴ m
  4. 7.28 × 10⁻¹⁰ m
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Answer: A

λ = hm × v = 6.63 × 10⁻³⁴9.1 × 10⁻³¹ × 2.0 × 10⁶ = 6.63 × 10⁻³⁴1.82 × 10⁻²⁴ = 3.64 × 10⁻¹⁰ m.

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