The work function of a metal is 3 eV . The color of the visible light that is required to cause emission of photoelectrons is

Solution

Physics sir · 2 steps

Turn the 3 eV work function into a threshold wavelength, then see which colour is short enough.

1. Condition for emission. A photon frees an electron only if its energy hν is at least the work function φ. Higher energy means shorter wavelength, so there is a longest wavelength that still works.

   λ₀ = hc/φ = (1240 eV nm)/(3 eV) ≈ 413 nm

   Colour	Typical wavelength	Photon energy 1240/λ	At least 3 eV?
   Red	about 700 nm	about 1.8 eV	No
   Yellow	about 580 nm	about 2.1 eV	No
   Green	about 530 nm	about 2.3 eV	No
   Blue (short-wavelength end)	towards 400 nm	up to about 3.1 eV	Yes

2. So. Only light from the short-wavelength (blue) end of the visible spectrum, below about 413 nm, carries 3 eV or more per photon. The answer is (c), blue.

The idea to keep. Find the threshold wavelength with λ₀ = 1240/φ (in nm, with φ in eV). Only light of shorter wavelength than λ₀ can eject electrons, however bright or dim it is.

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