Worked Solutions
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.
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.