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 (work function: 4.3 eV)

The equation for the photoelectric effect is:

$E=\phi + K$ (1)

where

$E=\frac{hc}{\lambda}$ is the energy of the incident photon, with

h = Planck constant

c = speed of light

$\lambda$ = wavelength

$\phi$ = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

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

$eV=K$

So we can rewrite (1) as

$E=\phi + eV$

where we have:

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

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

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

Converting into electronvolts,

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

And now we can solve eq.(1) to find the work function of the metal:

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

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so $\phi$ is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.