Answer to Question #180991 in Physics for Reina

Question

The work function of a metal is 3.6eV and is connected as an emitter in a photoelectric experiment where light with wavelength 732nm strikes metal. How to find if there is a minimum potential difference (reversed potential difference) that can be applied to ensure that there is no current in the circuit due to the movement of photoelectrons to the collector?

Answer 1

This reversed potential difference should be chosen big enough to suppress the kinetic energy of the emitted photoelectrons to ensure that there is no current in the circuit.

According to the Einstein's formula (see https://en.wikipedia.org/wiki/Photoelectric_effect#Theoretical_explanation) the kinetic energy of the released electrons will be:

$$K = h\nu - W$$

where $W = 3.6eV$ is the work function of the metal, $h = 4,1\times 10^{-15}eV\cdot s$ is the Planck constant, and $\nu$ is the frequency of the light. The frequency is connected wiht wavelength $\lambda = 732nm = 7.32\times 10^{-7}m$ as follows:

$$\nu = \dfrac{c}{\lambda}$$

where $c = 3\times 10^{8}m/s$ is the speed of light. Subsittuting this into the expression for $K$, obtain:

$$K = \dfrac{hc}{\lambda} - W$$

In order to completely reduce this kinetic energy one should apply the following potential difference (according to the energy-work theorem):

$$V = K$$

if $K$ is measured in eV. Then $V$ will be in volts. Thus, obtain:

$$V = \dfrac{4,1\times 10^{-15}\cdot 3\times 10^8}{7.32\times 10^{-7}} - 3.6 \approx -1.9V$$

Since $\dfrac{hc}{\lambda} <W$, the photoelectrons will not be emitted (there is no enough energy in the light to overcome the work function of the metal). Thus, the minimum potential difference can be 0.

Answer. 0V.