The magnetic field associated with a light wave is given, at the origin, by $B = B_0\left[\sin(3.14\times10^7)ct + \sin(6.28\times10^7)ct\right]$. If this light falls on a silver plate having a work function of $4.7\ \text{eV}$, what will be the maximum kinetic energy of the photoelectrons? $(c = 3\times10^8\ \text{m s}^{-1},\ h = 6.6\times10^{-34}\ \text{J s})$
Answer: (B) $7.72\ \text{eV}$
The higher-frequency component gives the fastest electrons. Its angular frequency is
$$\omega = 6.28\times10^7\times3\times10^8 = 1.884\times10^{16}\ \text{rad/s},\qquad f = \frac{\omega}{2\pi} = 3\times10^{15}\ \text{Hz}$$
$$E = hf = 6.6\times10^{-34}\times3\times10^{15} = 1.98\times10^{-18}\ \text{J} \approx 12.4\ \text{eV}$$
$$K_{max} = 12.4 - 4.7 \approx 7.7\ \text{eV}$$
The closest option is $7.72\ \text{eV}$.
Solution by Sreeraj P, M.Sc Physics