Question

A photosensitive metallic surface has work function $h{f}_0$. If photons of energy 2$h{f}_0$ fall on this surface, the electrons come out with a maximum velocity $4\times {10}^{6}m{s}^{-1}$. When the photon energy is increased to $5h{f}_0$, then the maximum velocity is

• A. $2\times {10}^{7}m{s}^{-1}$
• B. $2\times {10}^{6}m{s}^{-1}$
• C. $8\times {10}^{6}m{s}^{-1}$
• D. $2\times {10}^{5}m{s}^{-1}$

Answer 1

The correct option is C $8\times {10}^{6}m{s}^{-1}$

Maximum velocity of electrons related to the work function is given by

$\frac{1}{2}m{v}_{max}^2=hf-h{f}_o$

$\frac{\frac{1}{2}m{v}_1^2}{\frac{1}{2}m{v}_2^2}=\frac{h{f}_1-h{f}_o}{h{f}_2-h{f}_o}$

${\left(\frac{v_1}{v_2}\right)}^2=\frac{h{f}_1-h{f}_o}{h{f}_2-h{f}_o}$

$\left(\frac{4}{v_2}\right)=\sqrt{\frac{2h{f}_o-h{f}_o}{5h{f}_o-h{f}_o}}=\sqrt{\frac{1}{4}}$

$\frac{4}{v_2}=\frac{1}{2}$

$v_2=8$
So, the answer is option (C).