Question

Energy required to stop the ejection of electrons from $Cu$ plate is 0.24 eV. When radiation of $\lambda$ = 253.7 nm strikes the plate. Calculate the work function.

Answer 1

Energy required to stop the ejection of electron $E_5=0.24ev$

$\lambda =253.7nm$

$\frac{hc}{\lambda }=E_s+Q_0$

$Q_0=\frac{hc}{\lambda }-E_s$

$Q_0=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{253.7\times {10}^{-5}\times 1.6\times {10}^{-15}}-0.24$

$=4.897-0.24=4.657$