Question

A photosensitive metallic surface is illuminated alternately with lights of wavelength 3100 A and 6200 A. It is observed that maximum speeds of the photoelectrons in two cases are in the ratio 2:1. The work function of the metal is ($\text{hc = 12400 eVA}$).
(A) 1 eV
(B) 2 eV
(C) $\dfrac{4}{3}\text{ eV}$
(D) $\dfrac{2}{3}\text{ eV}$

Answer 1

Photo-sensitivity is defined as the term which is used to describe the sensitivity to the ultraviolet from the sunlight and other light sources. The light sources can also include indoor fluorescent light and many others. This is the concept that is used to solve this question.

Complete step by step answer
Given that, a photosensitive metallic surface is illuminated alternatively with lights of wavelength 3100 A and 6200 A. Also, the maximum speeds of the photo-electrons in two given cases are observed to be 2: 1.
We know,
$K \cdot E=h \dfrac{c}{\lambda}-E_{o}$
$E_{0}$ is denoted as the work function of the metal.
Therefore, from the equation, we can determine that Kinetic Energy is directly proportional to the square of speed of the photo-electron.
Hence, the ratio of kinetic energy can be calculated as 4: 1.
$\therefore \dfrac{K\cdot {{E}_{A}}}{K\cdot {{E}_{B}}}=\dfrac{h\dfrac{c}{{{\lambda}_{A}}}-{{E}_{o}}}{h\dfrac{c}{{{\lambda }_{B}}}-{{E}_{o}}}$
$\Rightarrow \dfrac{4}{1}=\dfrac{\dfrac{12400}{3100}-{{E}_{0}}}{\dfrac{12400}{6200}-{{E}_{o}}}$
$\Rightarrow \dfrac{4}{1}=\dfrac{4-{{E}_{o}}}{2-{{E}_{o}}}$
$\therefore {{E}_{0}}=\dfrac{4}{3}\text{eV}$
Hence, we get the work function of the metal as $\dfrac{4}{3} eV$.

Therefore, the correct answer is Option C.

Note We know that when light energy of the sufficient intensity strikes on a surface of the material, some electrons of the material very close to the surface, gain sufficient energy to overcome the work function of the material and are emitted from the surface with the kinetic energy. These emitted electrons are called photo-electrons.