Question

Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively, is :

• A. $\sqrt{2}:1$
• B. $\sqrt{3}:1$
• C. $\sqrt{3}:\sqrt{2}$
• D. $1:\sqrt{2}$

Answer 1

The correct option is D $1:\sqrt{2}$

Maximum kinetic energy, ${K.E.}_{max}=incidentenergy-workfunction$

$\frac{1}{2}m{V}_1^2=2\varphi -\varphi$

$\frac{1}{2}m{V}_1^2=\varphi$

$V_1^2=\frac{2\varphi }{m}$

$V_1=\sqrt{\frac{2\varphi }{m}}$

and $\frac{1}{2}m{V}_2^2=3\varphi -\varphi$

$\frac{1}{2}m{V}_2^2=2\varphi$

$V_2=\sqrt{\frac{4\varphi }{m}}$
So, $\frac{V_1}{V_2}=\frac{\sqrt{\frac{2\varphi }{m}}}{\sqrt{\frac{4\varphi }{m}}}=\frac{1}{\sqrt{2}}$.

So, the answer is option (D).