Question

Photons of energies twice and five times the work function of a metal fall successively on the metal surface. What is the ratio of the maximum velocities of photoelectrons emitted in the two cases?

Answer 1

From Equation of photo-electric effect :-

K.E. of photoelectron $=\frac{1}{2}m{V}^2=hv-\varphi$

For 1st radiation :- $\frac{1}{2}m{V}^2=2\varphi -\varphi =\varphi$ __(1)

and Rer 2nd radiation :- $\frac{1}{2}m{V}_2^2=5\varphi -\varphi =4\varphi$ __(2)

From (1) and (2)

$\frac{v_1^2}{v_2}=\frac{1}{4}\Rightarrow \frac{v_1}{v_2}=\frac{1}{2}$