These questions revise the lessons on photons and the photoelectric effect, starting with explaining the photoelectric effect. Use h = 6.63 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹ and electron mass 9.1 × 10⁻³¹ kg.
Try each question on paper first.
Questions
Question 1
Find the energy of a photon of frequency 6.0 × 10¹⁴ Hz.
Answer
E = hf = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 4.0 × 10⁻¹⁹ J.
Question 2
Find the energy of a photon of wavelength 4.0 × 10⁻⁷ m.
Answer
E = hc ÷ λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) ÷ 4.0 × 10⁻⁷ = 1.989 × 10⁻²⁵ ÷ 4.0 × 10⁻⁷ = 5.0 × 10⁻¹⁹ J.
Question 3
A metal has work function 3.0 × 10⁻¹⁹ J. Find its threshold frequency.
Answer
f₀ = W ÷ h = 3.0 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 4.5 × 10¹⁴ Hz.
Question 4
Light of frequency 8.0 × 10¹⁴ Hz falls on the metal in Question 3. Find the maximum kinetic energy and the maximum speed of the electrons.
Answer
hf = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.3 × 10⁻¹⁹ J, which is greater than W, so emission occurs.
KEmax = 5.3 × 10⁻¹⁹ − 3.0 × 10⁻¹⁹ = 2.3 × 10⁻¹⁹ J.
v = √(2KE ÷ m) = √(2 × 2.3 × 10⁻¹⁹ ÷ 9.1 × 10⁻³¹) ≈ 7.1 × 10⁵ m s⁻¹.
Question 5
Bright red light of frequency 4.0 × 10¹⁴ Hz falls on the metal in Question 3, and no electrons are emitted. Explain why.
Answer
The photon energy is 6.63 × 10⁻³⁴ × 4.0 × 10¹⁴ = 2.7 × 10⁻¹⁹ J, which is less than W = 3.0 × 10⁻¹⁹ J.
One photon gives its energy to one electron, so no electron gets enough to escape. Making the light brighter adds photons but does not raise the energy of any one photon.
Question 6
Light above the threshold frequency produces a photocurrent. What happens to the current and to the maximum kinetic energy if the intensity is halved?
Answer
Half as many photons arrive each second, so half as many electrons are emitted each second and the current halves.
Each photon still has energy hf, so the maximum kinetic energy is unchanged.
Question 7
A graph of KEmax against frequency is a straight line through (4.0 × 10¹⁴ Hz, 0) and (8.0 × 10¹⁴ Hz, 2.65 × 10⁻¹⁹ J). Find h and the work function.
Answer
Gradient = 2.65 × 10⁻¹⁹ ÷ (8.0 × 10¹⁴ − 4.0 × 10¹⁴) = 2.65 × 10⁻¹⁹ ÷ 4.0 × 10¹⁴ = 6.6 × 10⁻³⁴ J s.
W = hf₀ = 6.6 × 10⁻³⁴ × 4.0 × 10¹⁴ = 2.6 × 10⁻¹⁹ J.
Question 8
A lamp emits 0.50 W of light of wavelength 5.0 × 10⁻⁷ m. How many photons leave the lamp each second?
Answer
Photon energy = hc ÷ λ = 1.989 × 10⁻²⁵ ÷ 5.0 × 10⁻⁷ = 3.98 × 10⁻¹⁹ J.
Photons per second = 0.50 ÷ 3.98 × 10⁻¹⁹ = 1.3 × 10¹⁸.
If you got these wrong
Questions 1 to 3 need relating photon energy, frequency and wavelength. Questions 4 and 5 need explaining the photoelectric effect.
Question 6 needs distinguishing photon explanations from classical expectations. Question 7 needs reading threshold-frequency graphs.
To work through your wrong answers with a teacher, see online one-to-one Physics tuition.