A metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V₀. If the same surface is illuminated with radiation of wavelength 2 λ. the stopping potential becomes Vₒ/4. The threshold wavelength for this metallic surface will be

Solution

Physics sir · 4 steps

Write Einstein's equation for both wavelengths, then scale the second by 4 so both equal eV₀.

1. Two equations. With work function φ = hc/λ₀, Einstein's equation gives eV₀ = hc/λ - φ for wavelength λ. For 2λ it gives eV₀/4 = hc/2λ - φ.
2. Match the left sides. Multiply the second equation by 4 to get eV₀ = 2hc/λ - 4φ. Both equations now have eV₀ on the left, so their right sides are equal.

   hc/λ - φ = 2hc/λ - 4φ ⇒ 3φ = hc/λ

3. Threshold wavelength. So φ = hc/3λ. Comparing this with φ = hc/λ₀ gives λ₀ = 3λ.
4. So. The threshold wavelength is 3λ, option (a). It is longer than 2λ, as it must be, because light of wavelength 2λ still ejects electrons.

The idea to keep. Two wavelengths with two stopping potentials give two equations. Eliminate eV₀, find φ, then use λ₀ = hc/φ. Check that λ₀ is longer than every wavelength that ejects electrons.

Answer from the NTA final answer key. Solution by PrepWiser.