E.2 Quantum physics (HL only)

Frequency $f=1.2\times 10^{15}$ Hz, work function $\varphi =3.6$ eV, Planck's constant $h=4.14\times 10^{-15}$ eV·s.

$E=hf=(4.14\times 10^{-15})(1.2\times 10^{15})=4.97\text{ eV}$

$E_{k,\text{max}}=hf-\varphi =4.97-3.6=1.37\text{ eV}$

Since $hf>\varphi$, electrons are emitted with a maximum kinetic energy of $1.37$ eV.

We need the maximum kinetic energy of emitted electrons, which requires checking if emission occurs and applying $E_{k,\text{max}}=hf-\varphi$.

The photon energy is $hf=4.97$ eV, which exceeds $\varphi =3.6$ eV, so electrons are emitted. Eliminate this option.

This is the work function itself, not the kinetic energy. The kinetic energy is the difference $hf-\varphi$, not equal to $\varphi$.

This is the total photon energy. Part of it is used to overcome the work function, so the kinetic energy must be less than $4.97$ eV.

$E_{k,\text{max}}=4.97-3.6=1.37$ eV is the only remaining option and is correct.

Starting from $e{V}_s=hf-\varphi$, rearrange to: $V_s=\frac{h}{e}f-\frac{\varphi }{e}$

This is a linear equation $V_s=mf+c$, where the gradient $m=h/e$ and the y-intercept $c=-\varphi /e$.

The gradient of the $V_s$–$f$ graph equals $h/e$. Multiplying the measured gradient by the elementary charge $e$ gives an experimental value of Planck's constant $h$.

The gradient gives $h/e$, enabling determination of Planck's constant.

From $V_s=\frac{h}{e}f-\frac{\varphi }{e}$, the graph has a gradient, y-intercept, and x-intercept each encoding a different quantity.

$\varphi /e$ appears as the magnitude of the y-intercept, not the gradient. The y-intercept is $-\varphi /e$.

The threshold frequency is found from the x-intercept of the graph (where $V_s=0$), not the gradient.

$E_{k,\text{max}}$ corresponds to the y-value at a specific frequency, not to the gradient of the graph.

Only $h/e$ is the gradient of the $V_s$–$f$ graph, making this the correct answer.