The magnetic field associated with a light wave is given at the origin by
Question:

The magnetic field associated with a light wave is given at the origin by

$\mathrm{B}=\mathrm{B}_{0}\left[\sin \left(3.14 \times 10^{7}\right) \mathrm{ct}+\sin \left(6.28 \times 10^{7}\right) \mathrm{ct}\right] .$

If this light falls on a silver plate having a work function of $4.7 \mathrm{eV}$, what will be the maximum kinetic energy of the photoelectrons?

$\left(\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$

1. (1) $6.82 \mathrm{eV}$

2. (2) $12.5 \mathrm{eV}$

3. (3) $8.52 \mathrm{eV}$

4. (4) $7.72 \mathrm{eV}$

Correct Option: 4

Solution:

(4) According to question, there are two EM waves with different frequency,

$\mathrm{B}_{1}=\mathrm{B}_{0} \sin \left(\pi \times 10^{7} \mathrm{c}\right) \mathrm{t}$

and $\mathrm{B}_{2}=\mathrm{B}_{0} \sin \left(2 \pi \times 10^{7} \mathrm{c}\right) \mathrm{t}$

To get maximum kinetic energy we take the photon with higher frequency

using, $\mathrm{B}=\mathrm{B}_{0} \sin \omega \mathrm{t}$ and $\omega=2 \pi \mathrm{v} \Rightarrow \mathrm{v}=\frac{\omega}{2 \pi}$

$B_{1}=B_{0} \sin \left(\pi \times 10^{7} c\right) t \Rightarrow v_{1}=\frac{10^{7}}{2} \times c$

$\mathrm{B}_{2}=\mathrm{B}_{0} \sin \left(2 \pi \times 10^{7} \mathrm{c}\right) \mathrm{t} \Rightarrow \mathrm{v}_{2}=10^{7} \mathrm{c}$

where $c$ is speed of light $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Clearly, $v_{2}>v_{1}$

so KE of photoelectron will be maximum for photon of higher energy.

$v_{2}=10^{7} \mathrm{c} \mathrm{Hz}$

$\mathrm{h} v=\phi+\mathrm{KE}_{\max }$

energy of photon

$\mathrm{E}_{\mathrm{ph}}=\mathrm{h} v=6.6 \times 10^{-34} \times 10^{7} \times 3 \times 10^{9}$

$\mathrm{E}_{\mathrm{ph}}=6.6 \times 3 \times 10^{-19} \mathrm{~J}$

$=\frac{6.6 \times 3 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=12.375 \mathrm{eV}$

$\mathrm{KE}_{\max }=\mathrm{E}_{\mathrm{ph}}-\phi$

$=12.375-4.7=7.675 \mathrm{eV} \approx 7.7 \mathrm{eV}$