Question

Electromagnetic radiations having $\lambda =310\mathring{A}$ are subjected to a metal sheet having work function $12.8eV$. What will be the velocity of photoelectrons with maximum kinetic energy?

• A. $2.18\times {10}^{6}m/s$
• B. $2.18\sqrt{2}\times {10}^{6}m/s$
• C. $8.17\times {10}^{6}m/s$
• D. None of the above

Answer 1

Answer: (B)

$2.18\sqrt{2}\times {10}^{6}m/s$

Kinetic energy is given by the expression,

$K.E=\frac{hc}{{\lambda }_o}-\text{work function}$
$\frac{hc}{{\lambda }_o}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3.1\times {10}^{-}8}=64.1\times {10}^{-19}J$
Work function $=12.8eV=12.8\times 1.602\times {10}^{-19}=20.51\times {10}^{-19}J$
$K.E=64.1\times {10}^{-19}-20.51\times {10}^{-19}=43.59\times {10}^{-19}J$
$\frac{1}{2}m{v}^2=43.59\times {10}^{-19}$

and m =mass of electron $=9.1\times {10}^{-31}kg$
So, $v$ is calculated as $v=3.09\times {10}^6=2.18\sqrt{2}\times {10}^{6}m/s$.

Hence, the correct option is $B$