Question

The binding of electrons in a metal is $301kJmo{l}^{-1}$. The threshold frequency of the metal is $x\times {10}^{14}se{c}^{-1}$, where $x$ (approx.) is:

Answer 1

Binding energy of one electron$=\frac{301\times {10}^3}{6.02\times {10}^{23}}$$=5\times {10}^{-19}J$.
$B.E.=h{v}_0$.
$5\times {10}^{-19}=6.625\times {10}^{-34}\times v_0$.
$v_0=\frac{5\times {10}^{-19}}{6.625\times {10}^{-34}}$$=7.55\times {10}^{14}se{c}^{-1}$.