Question

The electric field of light wave is given as \[ E = 10^3 \cos \left( \frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t \right)\hat{j} \, \mathrm{N/C} \] This light falls on a metal plate of work function \( 1.5\, \mathrm{eV} \). The stopping potential of the photoelectron is \( \_\_\_ \, \mathrm{V} \).
(Energy of photon = \( \frac{1240}{\lambda \, (\text{in nm})} \, \mathrm{eV} \))

Hint:

Stopping potential (in volts) = max kinetic energy (in eV). No numerical conversion factor needed!

Answer 1

Step 1: Understanding the Concept:
Einstein’s photoelectric equation relates the energy of an incident photon to the work function of the metal and the maximum kinetic energy of the emitted photoelectrons. The stopping potential is numerically equal to the maximum kinetic energy in electron-volts.
Step 2: Key Formula or Approach:
1. Standard wave equation: \( E = E_0 \cos(kx - \omega t) \).
2. Wavelength \( \lambda = \frac{2\pi}{k} \).
3. Einstein equation: \( E_{\text{photon}} = \Phi + K_{\max} = \Phi + e V_s \).
Step 3: Detailed Explanation:
From the given equation, the wave number \( k = \frac{2\pi}{5 \times 10^{-7}} \).
Thus, the wavelength \( \lambda = 5 \times 10^{-7} \text{ m} = 500 \text{ nm} \).
Energy of the incident photon:
\[ E_{\text{photon}} = \frac{1240}{500} \text{ eV} = 2.48 \text{ eV} \]
Given work function \( \Phi = 1.5 \text{ eV} \).
Using the photoelectric equation:
\[ 2.48 \text{ eV} = 1.5 \text{ eV} + K_{\max} \]
\[ K_{\max} = 2.48 - 1.5 = 0.98 \text{ eV} \]
The stopping potential \( V_s \) in Volts is equal to the maximum kinetic energy in eV:
\[ V_s = 0.98 \text{ V} \]
Step 4: Final Answer:
The stopping potential is 0.98 V.