A photocathode of work function 1.5 eV is illuminated by a monochromatic source of fixed optical power. Initially the wavelength is 600 nm. It is then reduced to 300 nm, while the optical power and quantum efficiency remain unchanged. Which statement is correct?
Options

Question

A photocathode of work function 1.5 eV is illuminated by a monochromatic source of fixed optical power. Initially the wavelength is 600 nm. It is then reduced to 300 nm, while the optical power and quantum efficiency remain unchanged. Which statement is correct?

Options

G1 Physics · Accepted Answer

At fixed optical power P, photon rate is Ṅγ = P ÷ Eγ. Since Eγ = hc ÷ λ, we have Ṅγ ∝ λ. Reducing the wavelength from 600 nm to 300 nm halves the photon rate, so with unchanged quantum efficiency the saturation current also halves. Now compare maximum kinetic energies. At 600 nm: Eγ = 1240 ÷ 600 ≈ 2.067 eV, so Kₘₐₓ ≈ 2.067−1.5 = 0.567 eV. At 300 nm: Eγ = 1240 ÷ 300 ≈ 4.133 eV, so Kₘₐₓ ≈ 2.633 eV. Therefore the stopping potential increases while the saturation current halves. Correct option: B.