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Quiz Chapter 7: Quantum Physics

10 questions · Form 5 Physics Bab 7: Quantum Physics

Question 1 of 10Score: 0

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)

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. Light of photon energy 5.0 × 10⁻¹⁹ J strikes a metal surface with a work function of 3.2 × 10⁻¹⁹ J. What is the maximum kinetic energy of the emitted photoelectrons?

  1. 8.2 × 10⁻¹⁹ J
  2. 1.8 × 10⁻¹⁹ J
  3. 1.56 × 10⁻¹⁹ J
  4. 3.2 × 10⁻¹⁹ J
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Answer: B

K_max = E_photon - Work Function = 5.0 × 10⁻¹⁹ J - 3.2 × 10⁻¹⁹ J = 1.8 × 10⁻¹⁹ J.

3. 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.

4. 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.

5. Which change increases the NUMBER of photoelectrons emitted per second from a metal illuminated by light of frequency f > f₀?

  1. Increasing the intensity (brightness) of the light source
  2. Increasing the threshold frequency of the metal
  3. Decreasing the frequency of the incident light
  4. Decreasing the surface area of the metal
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Answer: A

Higher light intensity means a greater photon arrival rate per second, producing more emitted photoelectrons per second.

6. 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.

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. 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.

9. 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.

10. 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.

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