Question

The minimum energy required to overcome the attractive forces between an electron and the surface of $Ag$ metal is $552\times {10}^{-19}J$. What will be the maximum kinetic energy of electrons ejected out from $Ag$ which is being exposed to $UV$ light of $\lambda =360\mathring{A}$?

Answer 1

Energy of the photon absorbed $=\frac{h\cdot c}{\lambda }=\frac{6.625\times {10}^{-27}\times 3\times {10}^{10}}{360\times {10}^{-8}}$
$=5.52\times {10}^{-11}erg=5.52\times {10}^{-18}J$
$E\left(photon\right)=$ Work function $+KE$
$KE=5.52\times {10}^{-18}-5.52\times {10}^{-19}$
$=49.68\times {10}^{-19}J$.