Module 7: The Nature of Light — Worked Solutions (HSC Physics)
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Worked examples for HSC Physics Module 7: The Nature of Light. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Physics — Module 7: The Nature of Light. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Useful constants: Planck's constant $h = 6.63 \times 10^{-34}\ \text{J s}$, speed of light $c = 3.0 \times 10^8\ \text{m s}^{-1}$, and $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$.
Example 1 — The photoelectric effect
Question
Light of wavelength $4.0 \times 10^{-7}\ \text{m}$ is shone onto a metal surface with a work function of $2.0\ \text{eV}$. Using $h = 6.63 \times 10^{-34}\ \text{J s}$, $c = 3.0 \times 10^8\ \text{m s}^{-1}$ and $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$, calculate the maximum kinetic energy of the emitted photoelectrons in joules.
Solution
First find the photon energy from $E = \dfrac{hc}{\lambda}$.
$$E = \frac{(6.63\times10^{-34})(3.0\times10^8)}{4.0\times10^{-7}} = 4.97\times10^{-19}\ \text{J}$$
Convert the work function to joules: $\phi = 2.0 \times 1.6\times10^{-19} = 3.2\times10^{-19}\ \text{J}$.
Einstein's photoelectric equation gives the maximum kinetic energy:
$$K_{max} = E - \phi = 4.97\times10^{-19} - 3.2\times10^{-19} = 1.77\times10^{-19}\ \text{J}$$
So $K_{max} \approx 1.8\times10^{-19}\ \text{J}$. Convert the work function to joules before subtracting — mixing eV and J is the classic error here.
The photoelectric effect tells us that each photon delivers a fixed packet of energy. Some of that energy is used to free the electron from the metal (the work function $\phi$), and whatever is left becomes the electron's kinetic energy.
First, the photon's energy from its wavelength:
$$E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3.0\times10^8)}{4.0\times10^{-7}} = 4.97\times10^{-19}\ \text{J}$$
We must compare like with like, so convert the work function from electronvolts to joules: $\phi = 2.0 \times 1.6\times10^{-19} = 3.2\times10^{-19}\ \text{J}$.
Now apply Einstein's equation, which is really just energy conservation:
$$K_{max} = E - \phi = 4.97\times10^{-19} - 3.2\times10^{-19} = 1.77\times10^{-19}\ \text{J}$$
So the fastest electrons carry about $1.8\times10^{-19}\ \text{J}$. The reason there's a maximum is that surface electrons need the least energy to escape, so they keep the most.
Photon energy, then subtract work function.
- $E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times10^{-34})(3.0\times10^8)}{4.0\times10^{-7}} = 4.97\times10^{-19}\ \text{J}$
- $\phi = 2.0 \times 1.6\times10^{-19} = 3.2\times10^{-19}\ \text{J}$
- $K_{max} = E - \phi = 4.97\times10^{-19} - 3.2\times10^{-19}$
- $K_{max} = 1.77\times10^{-19}\ \text{J} \approx 1.8\times10^{-19}\ \text{J}$
Where the marks go
- 1 mark: Calculates photon energy $E = \dfrac{hc}{\lambda} = 4.97\times10^{-19}\ \text{J}$
- 1 mark: Converts work function to joules $\phi = 3.2\times10^{-19}\ \text{J}$
- 1 mark: Applies $K_{max} = E - \phi$
- 1 mark: Correct $K_{max} = 1.8\times10^{-19}\ \text{J}$ with units
Key idea
Einstein's photoelectric equation $K_{max} = \dfrac{hc}{\lambda} - \phi$ is energy conservation: photon energy in, work function to escape, kinetic energy out — keep all terms in joules.
Example 2 — Special relativity (time dilation)
Question
A spacecraft travels past Earth at $0.80c$, where $c = 3.0 \times 10^8\ \text{m s}^{-1}$. An astronaut on board measures a journey to take $5.0\ \text{years}$ of proper time. Calculate the time elapsed for an observer on Earth.
Solution
Use the time dilation formula $t = \dfrac{t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}$, where $t_0$ is the proper time measured on the spacecraft.
With $v = 0.80c$, $\dfrac{v^2}{c^2} = 0.64$, so:
$$\sqrt{1 - 0.64} = \sqrt{0.36} = 0.60$$
$$t = \frac{5.0}{0.60} = 8.3\ \text{years}$$
So the Earth observer measures $8.3\ \text{years}$. The proper time is always the shortest — the moving clock runs slow as seen from Earth.
Time dilation means a moving clock is observed to tick more slowly. The proper time $t_0$ is the time measured in the frame where the events happen at the same place — here, on the spacecraft, where the astronaut's own clock reads $5.0$ years.
The Earth observer sees this clock running slow, so they measure a longer time:
$$t = \frac{t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}$$
With $v = 0.80c$ we get $\dfrac{v^2}{c^2} = 0.80^2 = 0.64$, so the denominator is $\sqrt{1 - 0.64} = \sqrt{0.36} = 0.60$.
$$t = \frac{5.0}{0.60} = 8.3\ \text{years}$$
So Earth measures $8.3$ years while the astronaut ages only $5.0$ years. The reason is that nothing can exceed $c$, so the universe "stretches" time to keep the speed of light constant for everyone.
Time dilation: $t = \dfrac{t_0}{\sqrt{1 - v^2/c^2}}$, $t_0$ = proper time.
- $\dfrac{v^2}{c^2} = 0.80^2 = 0.64$
- $\sqrt{1 - 0.64} = \sqrt{0.36} = 0.60$
- $t = \dfrac{5.0}{0.60} = 8.3\ \text{years}$
Earth observer: $8.3\ \text{years}$.
Where the marks go
- 1 mark: Identifies $t_0 = 5.0\ \text{years}$ as proper time and states the time dilation formula
- 1 mark: Correctly evaluates the Lorentz factor denominator $\sqrt{1 - 0.64} = 0.60$
- 1 mark: Correct dilated time $t = 8.3\ \text{years}$
Key idea
Proper time $t_0$ is measured in the moving frame; the stationary observer measures a longer time $t = \dfrac{t_0}{\sqrt{1 - v^2/c^2}}$ — moving clocks run slow.
Frequently asked questions
Step-by-step solutions to exam-style questions on The Nature of Light in HSC Physics, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Physics syllabus for The Nature of Light, using the methods and notation expected in the exam.