Question

Light of wavelength $0.6\mu m~$ from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is $0.5V$. With light of wavelength $0.4\mu m~$from a mercury vapor lamp, the stopping potential is $1.5~V$. Then, the work function [in electron volts] of the photocell surface is
(A) $0.75eV$
(B) $1.5eV$
(C) $3eV$
(D) $2.5eV$

Answer 1

The photoelectric effect takes place as the emission of electrons when electromagnetic radiation, like light, hits a material. Electrons emitted in this way are called photoelectrons.
At stopping potential, no electrons reach the collector and there will be no current. It is independent of intensity.

Formula used: $KE=hf-\varphi$ where $KE$ is the kinetic energy, $h$ is the Planck's constant, $f$is the frequency and $\varphi$is the work function.
$hf$ is together considered as the energy of the photon.

Complete step by step answer:
The work function of the photocell surface is defined as the amount of energy that binds the electron to the metal. The electron that is knocked out of the metal gets energy from the photon.
We know that, $KE=hf-\varphi$
We’re given that stopping potential=0.5 V
Hence $KE=0.5eV$, since the maximum kinetic energy is the product of stopping potential and unit charge.
Energy of light with wavelength $0.6\mu m~$is $hf=\dfrac{hc}{\lambda }=\dfrac{6.626\text{ }\times \text{ }{{10}^{-34}}\times 3\times {{10}^{8}}}{0.6\times {{10}^{-6}}}=2.05eV$
$KE=hf-\varphi$
Hence, $\varphi =hf-KE$
$=2.05eV-0.5eV=1.5eV$

So, the correct answer is Option B.

Note:
The concept of photoelectric effect is important as classical physics was unable to explain the following
Why the electrons are emitted if the light frequency falls below fc
Why does maximum KE doesn’t depend on intensity?
Why are the electrons ejected almost instantaneously
And why does KE increases with increasing frequency