When UV light of wavelength 300 nm is incident on the metal surface having work function 2.13 eV, electron emission takes place. The stopping potential is :
(Given hc =1240 eV nm )

Solution

Physics sir · 3 steps

Find the photon's energy in eV, subtract the work function, and what is left is the stopping potential.

1. Photon energy. With hc = 1240 eV nm, the energy of one photon in eV is 1240 divided by the wavelength in nm. Here E = 1240/300 = 4.13 eV.

   eV₀ = Kₘₐₓ = hc/λ - φ

2. Subtract work function. The fastest electron keeps whatever energy is left after paying the work function: Kₘₐₓ = 4.13 - 2.13 = 2.0 eV.
3. So. A kinetic energy of 2 eV needs a retarding potential of 2 V to stop it, so V₀ = 2 V, option (b).

The idea to keep. Keep everything in eV. Then Kₘₐₓ in eV and the stopping potential in volts are the same number, and you never need to touch 1.6 × 10⁻¹⁹.

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