Radiations of wavelength 5000 Å fall on a metal plane whose work function is 1.9 eV. Calculate the stopping potential. [h = 6.63 × 10⁻³⁴ Js; c = 3 ×…

Explanation

This question is about the photoelectric effect. When light hits a metal surface, it can knock out electrons. The stopping potential is the voltage needed to stop the fastest electrons from reaching the other plate.

We use Einstein’s photoelectric equation: eV₀ = hf − φ, where V₀ is the stopping potential, h is Planck’s constant, f is the frequency of the light, and φ is the work function of the metal.

Step 1: Find the energy of the incoming photon.
E = hc/λ
λ = 5000 Å = 5000 × 10⁻¹⁰ m = 5 × 10⁻⁷ m
E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (5 × 10⁻⁷)
E = (19.89 × 10⁻²⁶) / (5 × 10⁻⁷)
E = 3.978 × 10⁻¹⁹ J

Step 2: Convert photon energy to eV.
E = 3.978 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 2.49 eV

Step 3: Find the stopping potential.
eV₀ = E − φ
eV₀ = 2.49 − 1.9 = 0.59 eV
Since eV₀ = eV₀ (the e cancels), V₀ = 0.59 V

The stopping potential is 0.59 V. This means a voltage of 0.59 V is enough to stop the fastest electrons emitted from the metal surface.