Question

If the electric field of EM wave is given by \( 60 \sin(3 \times 10^{14}t) + \sin(12 \times 10^{14}t) \) at \( x = 0 \) falls on a photo sensitive material having work function 2.8 eV. Find the maximum kinetic energy (in eV) of ejected electrons.

• A. 2.52 eV
• B. 2.16 eV
• C. 2.00 eV
• D. 2.34 eV

Hint:

The maximum kinetic energy of photoelectrons depends on the frequency of the incident light and the work function of the material. Use the highest frequency when dealing with a combination of waves.

Answer 1

The Correct Option is B

Step 1: Use the formula for maximum kinetic energy of photoelectrons.
The maximum kinetic energy of ejected electrons is given by the equation: \[ K_{\text{max}} = hf - \phi \] Where \( f \) is the frequency of the incident EM wave, and \( \phi \) is the work function.
Step 2: Find the frequency of the incident EM wave.
The electric field given in the problem is a combination of two waves with frequencies \( 3 \times 10^{14} \) and \( 12 \times 10^{14} \). The frequency corresponding to the highest energy will determine the maximum kinetic energy.
The frequency is \( f = 12 \times 10^{14} \, \text{Hz} \). The energy is given by: \[ E = h f = 6.626 \times 10^{-34} \times 12 \times 10^{14} = 4.963 \, \text{eV} \] Subtracting the work function \( \phi = 2.8 \, \text{eV} \), we get: \[ K_{\text{max}} = 4.963 - 2.8 = 2.16 \, \text{eV} \] Step 3: Conclusion.
The maximum kinetic energy of the ejected electrons is \( 2.16 \, \text{eV} \), which corresponds to option (2).