Question

A light wave described by E = 60[\(\sin(3 \times 10^{15})t + \sin(12 \times 10^{15})t\)] (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=\(6.6 \times 10^{-34}\) J.s. and e=\(1.6 \times 10^{-19}\) C)

• A. 3.8
• B. 5.1
• C. 6.0
• D. 7.8

Hint:

The photoelectric effect is a quantum phenomenon where one photon interacts with one electron.
If the incident light contains multiple frequencies, it's like having multiple types of photons.
The maximum kinetic energy of an ejected electron will always be determined by the highest energy (highest frequency) photon, as it provides the largest energy packet to the electron.

Answer 1

The Correct Option is B

To solve this problem, we need to determine the maximum kinetic energy of photoelectrons ejected from the metal surface when exposed to a light wave. The wave equation given is:

E = 60\left[\sin(3 \times 10^{15}t) + \sin(12 \times 10^{15}t)\right]

This equation represents the superposition of two light waves with angular frequencies \omega_1 = 3 \times 10^{15} \, \text{rad/s} and \omega_2 = 12 \times 10^{15} \, \text{rad/s}. The angular frequency \omega is related to the frequency f by:

\omega = 2\pi f

Therefore, the frequencies of these components are:

f_1 = \frac{3 \times 10^{15}}{2\pi} Hz and f_2 = \frac{12 \times 10^{15}}{2\pi} Hz.

The energies of photons corresponding to these frequencies are given by:

E_{photon} = h \cdot f

Using h = 6.6 \times 10^{-34} \, \text{J.s}, calculate the photon energies:

• E_{photon,1} = \frac{6.6 \times 10^{-34} \cdot 3 \times 10^{15}}{2\pi} \approx 3.15 \times 10^{-19} \, \text{J}
• E_{photon,2} = \frac{6.6 \times 10^{-34} \cdot 12 \times 10^{15}}{2\pi} \approx 1.26 \times 10^{-18} \, \text{J}

Convert these energies to electron volts (1 eV = 1.6 \times 10^{-19} \, \text{J}):

• E_{photon,1} \approx \frac{3.15 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.97 \, \text{eV}
• E_{photon,2} \approx \frac{1.26 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 7.88 \, \text{eV}

The maximum kinetic energy of ejected photoelectrons is given by the difference between the photon energy and the work function \phi of the metal (here, \phi = 2.8 \, \text{eV}):

• K_{max,1} = E_{photon,1} - \phi = 1.97 - 2.8 = -0.83 \, \text{eV} \, \text{(not enough energy to eject electron)}
• K_{max,2} = E_{photon,2} - \phi = 7.88 - 2.8 = 5.08 \, \text{eV}

The maximum kinetic energy of ejected photoelectrons is approximately 5.1 \, \text{eV}, which corresponds to the photons with frequency 12 \times 10^{15} \, \text{rad/s}.

Thus, the correct answer is 5.1.