Question

The threshold frequency of a metal is $6\times {10}^{14}{s}^{-1}$. Calculate the kinetic energy of an electron emitted when radiation of frequency $1.1\times {10}^{15}{s}^{-1}$ hits the metal.

• A. $3.31\times {10}^{-19}$ J
• B. $1.13\times {10}^{-18}$ J
• C. $7.12\times {10}^{-19}$ J
• D. None of These

Answer 1

The correct option is A $3.31\times {10}^{-19}$ J

As per Einstein's photo electric equation,
$K.E=E-W_o$
$K.E=hv-h{v}_o$,
$\text{where h is Planck's constant}=6.626\times {10}^{-34}Js$
$KE=h(\nu -{\nu }_o)=(6.62\times {10}^{-34})(1.1\times {10}^{15}-6\times {10}^{14})$
$KE=3.31\times {10}^{-19}$ J