Notes: When a piece of metal is illuminated by a monochromatic light of wavelength lambda, then stopping potential is 3Vs. When the same surface is...

Thursday, July 14, 2016

When a piece of metal is illuminated by a monochromatic light of wavelength lambda, then stopping potential is 3Vs. When the same surface is...

The stopping potential refers to the potential difference required to stop an electron emitted from a metal after it is illuminated by the light. This potential difference is equal to the kinetic energy with which electron leaves the metal:

$\displaystyle{K}={h}{f{-}}\phi$ .

This is called photoelectric equation. Here, h is the Plank's constant, f is the frequency of the incident light and $\displaystyle\phi$ is the work function of the metal. The work function is related to the threshold frequency of the incident light: $\displaystyle\phi={h}{f}_{{0}}$ , where $\displaystyle{f}_{{0}}$ is the minimum frequency for which any electrons will be emitted.

The frequency of light can be expressed through its wavelength $\displaystyle\lambda$ as

$\displaystyle{f{=}}\frac{{c}}{\lambda}$ , where c is the speed of light.

In the given problem, the stopping potential for the light with wavelength $\displaystyle\lambda$

is 3V. The photoelectric equation becomes

$\displaystyle{3}{V}=\frac{{{h}{c}}}{\lambda}-\phi$

The stopping potential for the light with wavelength $\displaystyle{2}\lambda$ is V:

$\displaystyle{V}=\frac{{{h}{c}}}{{{2}\lambda}}-\phi$

These two equations can be solved together for the work function. Multiplying the second equation by -2 and adding it to the first equation results in

$\displaystyle{V}=\phi$

The work function equals $\displaystyle\phi=\frac{{{h}{c}}}{\lambda_{{0}}}={V}$ , where $\displaystyle\lambda_{{0}}$ is the threshold wavelength. Combining this with second of the photoelectric equations, we get