Photoelectric Effect

Part a

3 minutes 2 marks

Question

Show that the energy of an incident photon is approximately 2.8 eV. [2]

Hint Hint

Hint

Use the formula \\(E = hc/\\lambda\\) and convert the result from Joules to electronvolts using \\(1 \u00A0eV = 1.60 \u00A0×10^{-19} \u00A0J\\).

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E = hc/λ = (6.63e-34 × 3e8) / 450e-9 J ≈ 4.42e-19 J. E (eV) ≈ 4.42e-19 / 1.60e-19 ≈ 2.76 eV ≈ 2.8 eV.

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Part b

2 minutes 1 marks

Question

Calculate the maximum kinetic energy, in eV, of the emitted photoelectrons. [1]

Hint Hint

Hint

Use Einstein’s photoelectric equation: \\(E_{k,max} = E_{photon} – \\Phi\\). Use E_photon from part (a).

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E_k,max = E_photon – Φ = 2.76 eV – 2.1 eV = 0.66 eV.

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Part c(i)

2 minutes 2 marks

Question

The intensity of the incident light is increased, while the wavelength is kept constant. State and explain the effect, if any, on:
(i) the maximum kinetic energy of the photoelectrons. [2]

Hint Hint

Hint

Photon energy depends on wavelength, not intensity. E_{k,max} = E_photon – Φ.

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Unchanged. Max KE depends on photon energy (fixed λ) and Φ, not on intensity.

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Part c(ii)

2 minutes 2 marks

Question

(ii) the rate of emission of photoelectrons. [2]

Hint Hint

Hint

Higher intensity → more photons per second → more electrons per second.

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Increases. More photons → more electrons ejected per unit time.

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Part d

6 minutes 3 marks

Question

Explain how the photoelectric effect provides evidence for the particle nature of light, referencing specific experimental observations that contradict the wave model. [3]

Hint Hint

Hint

Threshold frequency, instantaneous emission, KE vs intensity/frequency.

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Evidence includes: 1. Threshold frequency: emission only above f₀, regardless of intensity. 2. Instantaneous emission: no measurable delay above f₀. 3. Max KE depends on frequency, not intensity. These support photons (E=hf).

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