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

Concept: Photoelectric equation: \[ K_{\max} = E - \phi \] Stopping potential: \[ V_0 = \frac{K_{\max}}{e} \ \text{(in volts when } K_{\max} \text{ is in eV)} \]

Step 1: Find wavelength.
Comparing with standard wave equation \(E = E_0 \cos(kx - \omega t)\): \[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{5 \times 10^{-7}} \Rightarrow \lambda = 5 \times 10^{-7} \, \text{m} = 500 \, \text{nm} \]

Step 2: Photon energy.
\[ E = \frac{1240}{\lambda(\text{in nm})} = \frac{1240}{500} = 2.48 \, \text{eV} \]

Step 3: Maximum kinetic energy.
\[ K_{\max} = E - \phi = 2.48 - 1.5 = 0.98 \approx 1.0 \, \text{eV} \]

Step 4: Stopping potential.
Since \(K_{\max} = eV_0\): \[ V_0 = \frac{K_{\max}}{e} = 1.0 \, \text{V} \]