Question

The work functions of two metals are $2.75eV$ and $2eV$ respectively. If these are irradiated by photons of energy $3eV$, the ratio of maximum momenta of the photoelectrons emitted respectively by them is

• A. 1 : 2
• B. 1 : 3
• C. 1 : 4
• D. 2 : 1
• E. 4 : 1

Answer 1

The correct option is A 1 : 2

maximum momenta

$p=\sqrt{2mE}$

$\frac{p_1}{p_2}=\sqrt{\frac{h\nu -{\varphi }_1}{h\nu -{\varphi }_2}}=\sqrt{\frac{3-2.75}{3-2}}=\sqrt{0.25}=1:2$