Question

A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the cathode.
a.) what metal is the cathode made?
b.) what is the stopping potential if the intensity of light is doubled?

Answer 1

a) zinc

The equation of the photoelectric effect is:

$E=\phi + K$ (1)

where

E is the energy of the incident light

$\phi$ is the work function

$K$ is the maximum kinetic energy of the emitted photoelectrons

Here the wavelength of the incident light is

$\lambda=200 nm = 2\cdot 10^{-7} m$

so the energy of the light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7} m}=9.95\cdot 10^{-19} J$

Converting into electronvolts,

$E=\frac{9.95\cdot 10^{-19}}{1.6\cdot 10^{-19} J/eV}=6.22 eV$ (2)

The stopping potential is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the electrical potential energy corresponding to the stopping potential (V=1.93 V) must be equal to the maximum kinetic energy of the photoelectrons,

$U=q V = K$

and since the charge of the electron is

1 q = 1 e

We have

$K=(1 e)(1.93 V)=1.93 eV$ (3)

Combining (1), (2) and (3), we find the work function of the material:

$\phi = E-K=6.22 eV-1.93 eV=4.29 eV$

So, the cathode is most likely made of zinc, which has a work function of 4.3 eV.

b) The stopping potential does not change

As we said in part A), the stopping potential is proportional to the maximum kinetic energy of the photoelectrons, K.

The intensity of light is proportional to the number of photons that hit the surface of the metal. However, the energy of these photons does not depend on the intensity, but only on the frequency of the light.

Therefore, the energy of the photons (E) does not change when the intensity of light is doubled. Also, the work function $\phi$ does not change: this means that the maximum kinetic energy of the photoelectrons, K, does not change, and so the stopping potential remains the same.