The magnetic field associated with a light wave is given, at the origin, by $B = B_0 \left[\sin \right(3.14 \times {10}^7)ct + \sin (6.28 \times {10}^7\left)ct\right]$. If this light falls on a silver plate having a work function of $4.7 eV,$ what will be the maximum kinetic energy of the photo electrons ? $(c = 3 \times l{0}^{8}m{s}^{–1}, h = 6.6 \times {10}^{–34} J-s)$

Given data:

The magnetic field associated with a light wave is given, at the origin, by $B = B_0 \left[\sin \right(3.14 \times {10}^7)ct + \sin (6.28 \times {10}^7\left)ct\right]$

Detailed solution:

$B=B_0\sin \left(\pi \times {10}^7\mathrm{C}\right)\mathrm{t}+{\mathrm{B}}_0\sin \left(2\pi \times {10}^7\mathrm{C}\right)\mathrm{t}$
since there are two$\mathrm{EM}$ waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency
$B_1=B_0\sin \left(\pi \times {10}^7\mathrm{C}\right)t\Rightarrow {\nu }_1=\frac{{10}^7}{2}\mathrm{C}$

${\mathrm{B}}_2=B_0\sin \left(2\pi \times {10}^7\mathrm{C}\right)t\Rightarrow {\nu }_2={10}^7\mathrm{C}$
Where $c$ is speed of light  $\mathrm{c}=3\times {10}^8 \mathrm{m}/s\Rightarrow {\nu }_2>{\mathrm{\nu }}_1$
so $\mathrm{KE}$ of photoelectron will be maximum for photon of higher energy. $v_2={10}^{7}c Hz$

As we know $\mathrm{hv}=\varphi +{\mathrm{KE}}_{\max }$
energy of photon
$E_{ph}=hv=6.6\times {10}^{-34}\times {10}^7\times 3\times {10}^8$

$$\begin{gathered} \Rightarrow {\mathrm{E}}_{\mathrm{ph}}=6.6\times 3\times {10}^{-19} \mathrm{J} \\ =\frac{6.6\times 3\times {10}^{-19}}{1.6\times {10}^{-19}}\mathrm{eV}=12.375\mathrm{eV} \end{gathered}$$

$$\begin{gathered} {\mathrm{KE}}_{\max }=E_{\mathrm{ph}}-\varphi \\ =12.375-4.7=7.675\mathrm{eV}\approx 7.72\mathrm{eV} \end{gathered}$$

Hence , the correct option is A