If the work function of a metal is 6.63 eV, then find its threshold frequency for photoelectric effect.

Answer :

Solution :

${\mathrm{\varphi <!--\; \varphi \; -->}}_0=h{v}_0\mathrm{}$

$6.63\times <!--\; \times \; -->1.6\times <!--\; \times \; -->{10}^{-<!--\; -\; -->19}=6.63\times <!--\; \times \; -->{10}^{-<!--\; -\; -->34}{}_{}$

$v_0$ $v_0=\frac{1.6\times <!--\; \times \; -->{10}^{-<!--\; -\; -->19}}{{10}^{-<!--\; -\; -->34}}$

$v_0=1.6\times <!--\; \times \; -->{10}^{15}Hz$