Question

The work function of a metal is 4.2eV. If radiation of $2000A^o$ falls on the metal, the kinetic energy of the fastest photoelectron is:

• A. $1.6\times {10}^{-19}J$
• B. $16\times {10}^{-10}J$
• C. $3.2\times {10}^{-19}J$
• D. $6.4\times {10}^{-10}J$

Answer 1

Answer: (C)

$3.2\times {10}^{-19}J$

Kinetic energy of a photon is given by the expression, $K.E.=\frac{hc}{{\lambda }_o}-workfunction$.
Fastest photon travels with the speed of light, i.e., $c=3\times {10}^{8}m{s}^{-1}$.
$\frac{hc}{{\lambda }_o}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{2\times {10}^{-7}}=9.93\times {10}^{-19}J$
Work function $=4.2eV=4.2\times 1.602\times {10}^{-19}=6.73\times {10}^{-19}J$
So, $K.E.=9.93\times {10}^{-19}-6.73\times {10}^{-19}=3.2\times {10}^{-19}J$