The magnetic field associated with a light wave is given, at the origin, by $$B = B_0 [\sin(3.14 \times 10^7)ct + \sin(6.28 \times 10^7)ct]$$. 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 photoelectrons? ($$c = 3 \times 10^8$$ m s$$^{-1}$$, $$h = 6.6 \times 10^{-34}$$ J s)

We are told that, at the origin, the magnetic field of the incident light is

$$B = B_0\Bigl[\sin\!\bigl(3.14\times10^{7}\,c\,t\bigr)+\sin\!\bigl(6.28\times10^{7}\,c\,t\bigr)\Bigr].$$

The arguments of the sine functions must be dimensionless, so each term contains an angular frequency $$\omega$$ multiplied by the time $$t$$. Comparing, we identify the two angular frequencies

$$\omega_1 = 3.14\times10^{7}\,c,\qquad \omega_2 = 6.28\times10^{7}\,c.$$

With the speed of light $$c = 3\times10^{8}\,\text{m\,s}^{-1}$$, we evaluate

$$\omega_1 = (3.14\times10^{7})(3\times10^{8}) = 9.42\times10^{15}\,\text{rad\,s}^{-1},$$

$$\omega_2 = (6.28\times10^{7})(3\times10^{8}) = 1.884\times10^{16}\,\text{rad\,s}^{-1}.$$

The ordinary linear frequencies $$f$$ are obtained from the relation $$f=\dfrac{\omega}{2\pi}$$. Thus

$$f_1 = \dfrac{\omega_1}{2\pi} = \dfrac{9.42\times10^{15}}{6.283} \approx 1.50\times10^{15}\,\text{Hz},$$

$$f_2 = \dfrac{\omega_2}{2\pi} = \dfrac{1.884\times10^{16}}{6.283} \approx 3.00\times10^{15}\,\text{Hz}.$$

For each frequency, the photon energy is given by the Planck relation $$E = h f$$, where $$h = 6.6\times10^{-34}\,\text{J\,s}$$.

For the first component,

$$E_1 = h f_1 = (6.6\times10^{-34})(1.50\times10^{15}) = 9.9\times10^{-19}\,\text{J}.$$

To convert joules to electron-volts we divide by $$1.6\times10^{-19}\,\text{J/eV}$$:

$$E_1 = \dfrac{9.9\times10^{-19}}{1.6\times10^{-19}} \approx 6.19\,\text{eV}.$$

For the second component,

$$E_2 = h f_2 = (6.6\times10^{-34})(3.00\times10^{15}) = 1.98\times10^{-18}\,\text{J},$$

$$E_2 = \dfrac{1.98\times10^{-18}}{1.6\times10^{-19}} \approx 12.38\,\text{eV}.$$

The silver plate has a work function $$\phi = 4.7\,\text{eV}$$. According to the photoelectric equation

$$K_{\max} = h f - \phi,$$

the maximum kinetic energies corresponding to the two frequencies are

$$K_{\max}^{(1)} = E_1 - \phi = 6.19\,\text{eV} - 4.7\,\text{eV} = 1.49\,\text{eV},$$

$$K_{\max}^{(2)} = E_2 - \phi = 12.38\,\text{eV} - 4.7\,\text{eV} = 7.68\,\text{eV}.$$

The incident light actually contains both components, and the photoelectrons can originate from either. The maximum kinetic energy will obviously come from the higher-frequency (higher-energy) photons, so

$$K_{\max} \approx 7.68\,\text{eV} \;(\text{rounded to }7.72\,\text{eV}).$$

Hence, the correct answer is Option B.