A Level Physics 9702: Stopping Potential vs Frequency Graph (Photoelectric Effect)

In an experiment on the photoelectric effect, light of various frequencies $f$ is shone on a metal photocathode, and the stopping potential $V_s$ needed to reduce the photocurrent to zero is measured for each frequency. A graph of $V_s$ (on the $y$ -axis) against $f$ (on the $x$ -axis) is a straight line, with gradient $4.14\times10^{-15}\text{ V s}$ and a $y$ -intercept of $-1.90\text{ V}$ .

(a) Starting from Einstein's photoelectric equation, $hf=\phi+\tfrac12mv_{max}^2$ , and the definition of stopping potential, show that $V_s=\left(\dfrac{h}{e}\right)f-\dfrac{\phi}{e}$ . [2]

(b) Use the gradient of the graph to determine a value for the Planck constant $h$ . [2]

(c) Use the $y$ -intercept of the graph to determine the work function $\phi$ of the photocathode, giving your answer in both electron-volts and joules. [2]

(d) Calculate the threshold frequency $f_0$ of the photocathode. [2]

Part (a): Deriving the straight-line equation

The maximum kinetic energy of a photoelectron is stopped by a reverse potential difference $V_s$ , so $eV_s=\tfrac12mv_{max}^2$ . Substituting into Einstein’s photoelectric equation: $hf=\phi+\tfrac12mv_{max}^2 = \phi+eV_s$

Rearranging to make $V_s$ the subject, then dividing every term by $e$ : $eV_s = hf-\phi$ $V_s = \left(\frac{h}{e}\right)f-\frac{\phi}{e}$

This is of the form $y=mx+c$ with $y=V_s$ , $x=f$ , gradient $=h/e$ , and $y$ -intercept $=-\phi/e$ , as required.

Part (b): Finding the Planck constant

Comparing with $V_s=(h/e)f-\phi/e$ , the gradient of the graph equals $h/e$ : $h = (\text{gradient})\times e$

Substituting gradient $=4.14\times10^{-15}\text{ V s}$ and $e=1.60\times10^{-19}\text{ C}$ : $h = 4.14\times10^{-15}\times1.60\times10^{-19} = 6.624\times10^{-34}\text{ J s}$

Recompute as a check: $6.624\times10^{-34}/1.60\times10^{-19}=4.14\times10^{-15}\text{ V s}$ , matching the given gradient.

So $h=6.62\times10^{-34}\text{ J s}$ (3 s.f.), close to the accepted value.

Part (c): Finding the work function

Comparing with $V_s=(h/e)f-\phi/e$ , the $y$ -intercept equals $-\phi/e$ : $\phi = -(\text{intercept})\times e$

Substituting intercept $=-1.90\text{ V}$ : $\phi = -(-1.90)\times1.60\times10^{-19} = 1.90\times1.60\times10^{-19} = 3.04\times10^{-19}\text{ J}$

Since dividing by $e$ turns joules into electron-volts, this is equivalently: $\phi = 1.90\text{ eV}$

Recompute as a check: $3.04\times10^{-19}/1.60\times10^{-19}=1.90\text{ eV}$ , matching the magnitude of the given intercept.

Part (d): Threshold frequency

The threshold frequency $f_0$ is the frequency at which the stopping potential is exactly zero (a photon has just enough energy to release an electron with no kinetic energy left over). Setting $V_s=0$ in the straight-line equation: $0 = \left(\frac{h}{e}\right)f_0-\frac{\phi}{e} \implies f_0 = \frac{\phi/e}{h/e} = \frac{-(\text{intercept})}{\text{gradient}}$

Substituting the given values directly: $f_0 = \frac{1.90}{4.14\times10^{-15}} = 4.589\times10^{14}\text{ Hz}$

Recompute as a check, using the values of $h$ and $\phi$ found in parts (b) and (c): $f_0=\phi/h=3.04\times10^{-19}/6.624\times10^{-34}=4.589\times10^{14}\text{ Hz}$ , the same answer both ways.

So $f_0=4.59\times10^{14}\text{ Hz}$ (3 s.f.).

Final answers

(a) $V_s=(h/e)f-\phi/e$ , shown by substituting $eV_s=\tfrac12mv_{max}^2$ into Einstein’s equation and dividing by $e$ .
(b) $h=\boxed{6.62\times10^{-34}}\text{ J s}$
(c) $\phi=\boxed{1.90}\text{ eV} = \boxed{3.04\times10^{-19}}\text{ J}$
(d) $f_0=\boxed{4.59\times10^{14}}\text{ Hz}$