Question

When a photon of frequency $1.0\times {10}^{15}{s}^{-1}$ was allowed to hit a metal surface, an electron having $1.988\times {10}^{19}$J of kinetic energy was emitted. Calculate the threshold frequency of this metal.

Answer 1

We know that
$K.E.=hv-h{v}_0$
Given that
$V=1.0\times {10}^{15}{8}^{-1},K.E.=1.988\times {10}^{-19}$
$h{v}_0=hv-KE$
$V_0=\frac{hv}{h}-\frac{KE}{h}$
= $v-\frac{KE}{h}$
= ${10}^{15}-\frac{1.988\times {10}^{-19}}{6.626\times {10}^{-34}}$
$V_0={10}^{15}-0.3\times {10}^{15}$
= $0.7\times {10}^{15}$
$V_0=7\times {10}^{14}$