Question

The wavelength of incident light falling on a photosensitive surface is changed from 2000 Å to 2100 Å. The corresponding change in stopping potential is

• A. 0.03 V
• B. 0.3 V
• C. 3 V
• D. 3.3 V

Hint:

$hc/e = 12400$ eV·Å is a useful constant.

Answer 1

The Correct Option is B

Step 1: Photoelectric equation: $eVₛ = \frachcλ - φ$. Change: $eΔ Vₛ = hc\left(\frac1\lambda₁ - \frac1\lambda₂\right)$.

Step 2: $Δ Vₛ = \frachce\left(\frac12000×10^-10 - \frac12100×10^-10\right) = \frac124001.6×10^-19\left(\frac1002×10^-7\right)$? Using $hc/e = 12400$ eV·Å, $Δ Vₛ = 12400\left(\frac12000 - \frac12100\right) = 12400 × \frac1002000×2100 = 12400/42000 = 0.295$ V ≈ 0.3 V.