3.2.2.1
The E_k vs f graph gives work function, threshold frequency and Planck's constant
The Photoelectric Effect — AQA A-Level Physics
- Rearranging the photoelectric equation: $E_{k(\max)} = hf - \phi$.
- This has the form $y = mx + c$, so plotting $E_{k(\max)}$ against $f$ gives a straight line.
Reading the graph
- Gradient $= h$ (Planck's constant).
- $y$-intercept $= -\phi$ (negative work functionThe minimum energyThe capacity to do work. Measured in joules (J). required to liberate an electron from the surface of a metal.).
- $x$-intercept $= f_0$ (threshold frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz).The minimum frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). of incident radiation required to cause photoelectric emission from a particular metal surface.).
- No electrons are emitted for frequencies below $f_0$.
Worked Example
A graph of $E_k$ against $f$ for sodium has an $x$-intercept at $f_0 = 4.0 \times 10^{14}$ Hz. Calculate the work functionThe minimum energyThe capacity to do work. Measured in joules (J). required to liberate an electron from the surface of a metal. in eV.
Show Solution
1
Use the threshold condition
At $f_0$, $E_{k(\max)} = 0$, so $\phi = hf_0$.
2
Calculate in joules
$$\phi = (6.63 \times 10^{-34}) \times (4.0 \times 10^{14}) = 2.65 \times 10^{-19} \text{ J}$$
3
Convert to eV
$$\phi = \frac{2.65 \times 10^{-19}}{1.60 \times 10^{-19}} = 1.66 \text{ eV}$$
Answer
$\phi = 1.66$ eV
Worked Example
Show that the photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. energyThe capacity to do work. Measured in joules (J). of light with wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). 700 nm is about 1.8 eV.
Show Solution
1
Calculate in joules
$$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}) \times (3.0 \times 10^{8})}{700 \times 10^{-9}} = 2.84 \times 10^{-19} \text{ J}$$
2
Convert to eV
$$E = \frac{2.84 \times 10^{-19}}{1.60 \times 10^{-19}} = 1.78 \text{ eV} \approx 1.8 \text{ eV}$$
Answer
$E = 1.78$ eV $\approx 1.8$ eV
Examiner Tips and Tricks
- In 'show that' questions, always work to more significant figures than the target value, then state the answer rounds to the given value.
- Don't just plug numbers in and write the answer you were told.