When radiation of wavelength λ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is :

Solution

Physics sir · 4 steps

Write Einstein's equation for both wavelengths and subtract: the work function drops out at once.

1. Two equations. Einstein's equation hc/λ = φ + eV₀, in eV, gives hc/λ = φ + 4.8 for the first light. For the second light it gives hc/2λ = φ + 1.6.
2. Subtract them. Subtracting removes φ and leaves hc/λ - hc/2λ = 3.2, so hc/2λ = 3.2 eV. That makes hc/λ = 6.4 eV.
3. Work function. From the first equation, φ = 6.4 - 4.8 = 1.6 eV. This is one quarter of the photon energy at λ.

   λ₀ = hc/φ = 6.4/1.6 λ = 4λ

4. So. The threshold wavelength is 4λ, option (b).

The idea to keep. Two stopping potentials for the same metal means two Einstein equations. Subtract to kill φ, find hc/λ, then use λ₀ = hc/φ.

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