Question

If light of wavelength $412.5nm$ is incident on each of the metals given below, which ones will show photoelectric emission and why?

Metal	Work Function (eV)
$Na$	$1.92$
$K$	$2.15$
$Ca$	$3.20$
$Mo$	$4.17$

Answer 1

Wavelength of incident light $\lambda =412.5nm$
So, energy of the incident light $E=\frac{hc}{\lambda }=\frac{1240}{\lambda \left(\text{in nm}\right)}$ eV
$⟹E=\frac{1240}{412.5}=3$ eV

Photoelectric effect takes place only when the energy of the incident photon exceeds the work function of the metal.
Since, work function of $Na$ and $K$ is smaller than energy of incident light. So, only $Na$ and $K$ will show photoelectric emission.