A light wave described by $$E=60[\sin(3\times10^{15})t+\sin(12\times10^{15})t]$$ (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ___ eV. $$(h=6.6\times10^{-34}J.s\text{ and }e=1.6\times10^{19}C)$$

The incident electric field is a superposition of two monochromatic waves:

$$E = 60\,[\sin(3\times10^{15}t) + \sin(12\times10^{15}t)]$$

Hence the two angular frequencies are

$$\omega_1 = 3\times10^{15}\;{\rm s^{-1}},\qquad \omega_2 = 12\times10^{15}\;{\rm s^{-1}}$$

Step 1 - Convert angular frequency to ordinary frequency. The relation is

$$\nu = \frac{\omega}{2\pi}\qquad -(1)$$

Using $$(1)$$:

$$\nu_1 = \frac{3\times10^{15}}{2\pi}\;{\rm Hz} \approx \frac{3\times10^{15}}{6.283} \approx 4.78\times10^{14}\;{\rm Hz}$$

$$\nu_2 = \frac{12\times10^{15}}{2\pi}\;{\rm Hz} \approx \frac{12\times10^{15}}{6.283} \approx 1.91\times10^{15}\;{\rm Hz}$$

Step 2 - Photon energy for each component. The formula is

$$E_{\text{photon}} = h\nu \qquad -(2)$$

With $$h = 6.6\times10^{-34}\;{\rm J\,s}$$ we get

$$E_1 = 6.6\times10^{-34}\times4.78\times10^{14} \approx 3.15\times10^{-19}\;{\rm J}$$

$$E_2 = 6.6\times10^{-34}\times1.91\times10^{15} \approx 1.26\times10^{-18}\;{\rm J}$$

Step 3 - Convert these energies to electron-volts:

$$1\;{\rm eV} = 1.6\times10^{-19}\;{\rm J}$$

$$E_1 = \frac{3.15\times10^{-19}}{1.6\times10^{-19}} \approx 1.97\;{\rm eV}$$

$$E_2 = \frac{1.26\times10^{-18}}{1.6\times10^{-19}} \approx 7.88\;{\rm eV}$$

Step 4 - Decide which photons can eject electrons.

The work function of the metal is $$\phi = 2.8\;{\rm eV}$$.

• For $$E_1\;(1.97\;{\rm eV}) \lt \phi$$, photo-emission is impossible.
• For $$E_2\;(7.88\;{\rm eV}) \gt \phi$$, photo-emission occurs.

Step 5 - Maximum kinetic energy of the emitted electrons (Einstein’s equation):

$$K_{\max} = E_2 - \phi \qquad -(3)$$

$$K_{\max} = 7.88\;{\rm eV} - 2.8\;{\rm eV} = 5.08\;{\rm eV}$$

Rounded to one decimal place, $$K_{\max} \approx 5.1\;{\rm eV}$$.

Thus the maximum kinetic energy of the photo-electrons is 5.1 eV, which corresponds to Option D.