A light wave described by $E=60\left[\sin \left(3 imes 10^{15} ight) t+\sin \left(12 imes 10^{15} ight) t ight]$ (in SI units) falls on a metal surface of work function 2.8 eV . The maximum kinetic energy of ejected photoelectron is (approximately)

____ eV. $\left(h=6.6 imes 10^{-34} ight.$ J.s. and $\left.e=1.6 imes 10^{-19} \mathrm{C} ight)$

Question

A light wave described by $E=60\left[\sin \left(3 \times 10^{15}\right) t+\sin \left(12 \times 10^{15}\right) t\right]$ (in SI units) falls on a metal surface of work function 2.8 eV . The maximum kinetic energy of ejected photoelectron is (approximately)

____ eV. $\left(h=6.6 \times 10^{-34}\right.$ J.s. and $\left.e=1.6 \times 10^{-19} \mathrm{C}\right)$

____ eV. $\left(h=6.6 	imes 10^{-34}ight.$ J.s. and $\left.e=1.6 	imes 10^{-19} \mathrm{C}ight)$

Accepted Answer

D. 1. Einstein's photoelectric equation says maximum kinetic energy equals photon energy minus work function.

$$K_{\max} = E - \phi$$

2. Photon energy relates to angular frequency by $E = h
u = \dfrac{h\omega}{2\pi}$.

3. The given field has two angular frequencies mixed together: $\omega_1 = 3	imes10^{15}\ \mathrm{rad/s}$ and $\omega_2 = 12	imes10^{15}\ \mathrm{rad/s}$. Since higher frequency means higher photon energy, and $K_{\max}$ depends on the most energetic photons present, we use $\omega_{\max} = 12	imes10^{15}\ \mathrm{rad/s}$.

4. Converting this photon energy to eV:

$$E(	ext{in eV}) = \frac{h\cdot\omega_{\max}}{2\pi\cdot e} = \frac{(6.6	imes10^{-34})(12	imes10^{15})}{2	imes3.14	imes(1.6	imes10^{-19})} \approx 7.88\ \mathrm{eV}$$

5. Therefore:

$$K_{\max} = 7.88 - 2.8 = 5.08\ \mathrm{eV} \approx 5.1\ \mathrm{eV}$$

Hence, the answer is **Option D: 5.1 eV**.