Question

Let the energy of an emitted photoelectron be E and the wave-length of incident light be $\lambda$. What will be the change in E if $\lambda$ is doubled?
A.) E
B.) E/2
C.) 2E
D.) E/4

Answer 1

Hint: The energy of light wave incident will be utilized in detaching the electron from the surface and the remaining will be emitted as photoelectrons. So, we will first find out the energy of the light wave incident in both the cases given, and after comparing them, we will get to our results.

Formula used:
$\dfrac{hc}{\lambda}=\varphi + E$, where $\varphi$, where h is the Planck’s constant, c is the speed of the electromagnetic wave, $\varphi$ is the energy utilized in detaching the electron and E is the energy of photoelectron emitted.

Complete step by step solution
When an electromagnetic wave of certain wave-length $\lambda$ is incident on a metal surface, photoelectrons are emitted due to the absorption of incident radiation.

The energy of the wave with wavelength $\lambda$ is given by $\dfrac{hc}{\lambda}$, where h is the Planck’s constant and c is the speed of the wave.

When the light waves are incident, its energy gets utilized in detaching the electrons from its surface and then remaining energy is emitted in the form of photoelectrons.

So, we can write $\dfrac{hc}{\lambda}=\varphi + E$, where $\varphi$ is the energy utilized in detaching the electron and E is the energy of the photoelectron emitted. Here hc and $\varphi$ are constant, thus $\dfrac{1}{\lambda} \propto E$

Now, the initial energy of the photoelectron is E with wavelength $\lambda$, that means $E\propto \dfrac{1}{\lambda}$ ………. (i)

Assume that the energy of a photoelectron with wavelength $2\lambda$ be E’, that is $E’=\dfrac{1}{2\lambda}$ ………. (ii)

Taking ration of equations (i) and (ii), we get

$\dfrac{E}{E’}=\dfrac{2\lambda}{\lambda}\implies E’=\dfrac{E}{2}$

Hence, option b is the correct answer.

Note: An electron is detached from its surface only when its threshold frequency is reached due to the radiation falling on the surface. All the energy will not be converted to photoelectron energy but only the part that is left after the electron is detached.