Question

Radiations of two photon's energy, twice and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of photoelectrons emitted in two cases is:

• A. 1:2
• B. 1:3
• C. 1:4
• D. 1:1

Answer 1

The correct option is B 1:3

Photoelectric Law,

$K{E}_{max}=h\nu -W_o$

$\frac{1}{2}m{v}^2=h\nu -W_o$

$v_{max}=\sqrt{\frac{2(h\nu -W_o)}{m}}$

Given,

$h{\nu }_1=2W_o$ and

$h{\nu }_2=10W_o$

$\frac{v_1}{v_2}=\sqrt{\frac{h{\nu }_1-W_o}{h{\nu }_2-W_o}}$

$\frac{v_1}{v_2}=\sqrt{\frac{2W_o-W_o}{10W_o-W_o}}$

$\frac{v_1}{v_2}=\sqrt{\frac{1}{9}}$

$v_1:v_2=1:3$