Question

The threshold frequency $v_0$ for a metal is $7.0\times {10}^{14}{s}^{-1}.$ Calculate the kinetic energy of an electron emitted when radiation of frequency $v=1.0\times {10}^{15}{s}^{-1}$ hits the metal.

Answer 1

Given that,

$v_o=7.0\times {10}^{14}{s}^{-1}$

$v=1.0\times {10}^{15}{s}^{-1}$

As we know, $hv=h{v}_0+\frac{1}{2}m{v}^2$

i.e. $h(v-v_0)=K.E.$

$\therefore K.E.=(6.626\times {10}^{-34})\times (1.0\times {10}^{15}-7.0\times {10}^{14})=6.626\times {10}^{-34}\times {10}^{14}(10-7)$

$\therefore K.E=19.878\times {10}^{-20}J$