A metal has work function $2.5 \mathrm{eV}$. If a radiation of frequency $32 \times 10^{15} \mathrm{~Hz}$ is…
A metal has work function $2.5 \mathrm{eV}$. If a radiation of frequency $32 \times 10^{15} \mathrm{~Hz}$ is incident on this metal surface, then the maximum kinetic energy of ejected photoelectrons is (Planck's constant, $h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
$9.5 \mathrm{eV}$
$2.5 \mathrm{eV}$
$10.7 \mathrm{eV}$
$12.6 \mathrm{eV}$
Solution
Frequency of incident radiation,
$f=3.2 \times 10^{15} \mathrm{~Hz}$
Energy of incident photon $E, E=h f$
$\begin{aligned} & =6.6 \times 10^{-34} \times 3.2 \times 10^{15} \\ & =\frac{6.6 \times 10^{-34} \times 3.2 \times 10^{15}}{1.6 \times 10^{-19}} \mathrm{eV} \\ & =13.2 \mathrm{eV}\end{aligned}$
So, maximum K. E of emitted photoelectrons
$=E-\phi_0$
$=13.2-25$ $\left(\because \phi_0=2.5 \mathrm{eV}\right.$, given $)$
$=10.7 \mathrm{eV}$