The work functions for the following metals are given:
$Na: 2.75\;eV; K: 2.30\;eV; Mo: 4.17\;eV; Ni: 5.15\;eV$. Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\,\mathring{A}$ from a $He-Cd$ laser placed $1\;m$ away from the photocell? What happens if the laser is brought nearer and placed $50\;cm$ away?

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $\lambda = 3300\,\mathring{A} = 3300 \times 10^{-10}\,m$,$h = 6.63 \times 10^{-34}\,Js$,and $c = 3 \times 10^8\,m/s$.
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3300 \times 10^{-10}}\,J = 6.027 \times 10^{-19}\,J$.
Converting to electron-volts: $E = \frac{6.027 \times 10^{-19}}{1.6 \times 10^{-19}}\,eV \approx 3.77\,eV$.
Photoelectric emission occurs only if the incident photon energy $E$ is greater than the work function $\Phi$ of the metal.
Here,$E = 3.77\,eV$. Comparing this with the given work functions:
$Na (2.75\,eV) < 3.77\,eV$ (Emission occurs)
$K (2.30\,eV) < 3.77\,eV$ (Emission occurs)
$Mo (4.17\,eV) > 3.77\,eV$ (No emission)
$Ni (5.15\,eV) > 3.77\,eV$ (No emission)
Thus,$Mo$ and $Ni$ will not show photoelectric emission.
If the laser is moved to $50\;cm$,the intensity of the radiation increases,but the energy of individual photons remains $3.77\,eV$. Since the energy is still less than the work functions of $Mo$ and $Ni$,there will still be no photoelectric emission for these metals.