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Quiz Chapter 7: Quantum Physics

10 questions · Form 5 Physics Bab 7: Quantum Physics

10 questions · plays in fullscreen

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. Calculate the energy of a single photon of blue light with a frequency of 6.0 × 10¹⁴ Hz. (Planck's constant h = 6.63 × 10⁻³⁴ J s)

  1. 3.98 × 10⁻¹⁹ J
  2. 1.11 × 10⁻⁴⁷ J
  3. 3.98 × 10⁻²⁷ J
  4. 1.80 × 10⁻¹⁹ J
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Answer: A

E = h × f = (6.63 × 10⁻³⁴ J s) × (6.0 × 10¹⁴ Hz) = 3.978 × 10⁻¹⁹ J ≈ 3.98 × 10⁻¹⁹ J.

2. What is the gradient of a K_max against frequency (f) graph for any photoelectric metal surface?

  1. Work function (Φ)
  2. Planck's constant (h)
  3. Threshold frequency (f₀)
  4. Electron mass (m_e)
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Answer: B

Comparing K_max = hf - Φ to y = mx + c shows the gradient m equals Planck's constant h for all metals.

3. What happens to the energy of a photon when its wavelength increases?

  1. Increases proportionally
  2. Decreases because energy is inversely proportional to wavelength
  3. Remains unchanged
  4. Doubles automatically
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Answer: B

From E = hc/λ, photon energy is inversely proportional to wavelength λ.

4. 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.

5. An electron microscope offers significantly higher magnification and resolution than an optical light microscope because:

  1. Electrons have much shorter de Broglie wavelengths than visible light wavelengths
  2. Electrons travel at the speed of sound
  3. Electrons carry a positive charge that absorbs optical distortions
  4. Electrons expand glass lenses
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Answer: A

High-velocity electrons have extremely small de Broglie wavelengths (~pm), drastically reducing diffraction limits and allowing ultra-high resolution.

6. 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.

7. If the stopping potential for photoelectrons emitted from a metal plate is 2.0 V, what is the maximum velocity of the photoelectrons? (m_e = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)

  1. 8.39 × 10⁵ m s⁻¹
  2. 7.04 × 10¹¹ m s⁻¹
  3. 3.51 × 10⁵ m s⁻¹
  4. 1.20 × 10⁶ m s⁻¹
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Answer: A

K_max = e V_s = 1.6 × 10⁻¹⁹ × 2.0 = 3.2 × 10⁻¹⁹ J. Then 0.5 × m × v² = 3.2 × 10⁻¹⁹ => v = √[ 2 × 3.2 × 10⁻¹⁹9.1 × 10⁻³¹ ] = 8.39 × 10⁵ m s⁻¹.

8. 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.

9. What factor determines the MAXIMUM KINETIC ENERGY of photoelectrons emitted from a metal surface?

  1. Intensity of the incident light
  2. Frequency of the incident light
  3. Surface area of the illuminated metal plate
  4. Duration of light exposure
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Answer: B

From Einstein's equation K_max = hf - Work Function, kinetic energy depends strictly on photon frequency f.

10. If light of frequency f below the threshold frequency (f < f₀) shines on a metal surface, what happens?

  1. Photoelectrons are emitted very slowly after a long time delay
  2. No photoelectrons are emitted regardless of light intensity
  3. Photoelectrons are emitted with maximum kinetic energy
  4. The metal surface instantly melts
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Answer: B

If f < f₀, individual photons lack sufficient energy to overcome the work function, so zero photoelectrons are emitted regardless of brightness.

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