10 questions · Form 5 Physics Bab 7: Quantum Physics
10 questions · plays in fullscreen
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)
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. Why could classical wave theory NOT explain the photoelectric effect?
Answer: A
Wave theory assumed energy spread continuously over wave fronts, predicting time delays for dim light, whereas experiments proved emission is instantaneous.
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)
Answer: A
K_max = e × V_s = (1.6 × 10⁻¹⁹ C) × 1.5 V = 2.4 × 10⁻¹⁹ J.
4. What is the threshold frequency (f₀) of a metal?
Answer: B
Threshold frequency (f₀) is the minimum photon frequency needed to overcome the metal's work function (Φ = hf₀).
5. A solar panel in a satellite converts light energy directly into electrical energy using the:
Answer: B
Solar panels utilize photovoltaic solar cells operating on the photoelectric effect to convert solar photons directly into electric current.
6. 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?
Answer: B
K_max = E_photon - Work Function = 5.0 × 10⁻¹⁹ J - 3.2 × 10⁻¹⁹ J = 1.8 × 10⁻¹⁹ J.
7. What happens to the stopping potential (V_s) if light of a higher frequency is used in a photoelectric setup?
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.
8. 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)
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.
9. If light of frequency f below the threshold frequency (f < f₀) shines on a metal surface, what happens?
Answer: B
If f < f₀, individual photons lack sufficient energy to overcome the work function, so zero photoelectrons are emitted regardless of brightness.
10. What is the gradient of a K_max against frequency (f) graph for any photoelectric metal surface?
Answer: B
Comparing K_max = hf - Φ to y = mx + c shows the gradient m equals Planck's constant h for all metals.