The kinetic energy of the photoelectrons increases by 0.52 eV when the wavelength of incident light is changed from 500 nm to another wavelength which is approximately

Question

The kinetic energy of the photoelectrons increases by 0.52 eV when the wavelength of incident light is changed from 500 nm to another wavelength which is approximately (A) 700 nm (B) 400 nm (C) 1250 nm (D) 1000 nm

Tardigrade Admin · Accepted Answer

Here, change in kinetic energy, Δ K = 0.52 eV, λ = 500 nm, λ2 = ? We know , K1 = (hc/λ1 ) = φ K2 = (hc/λ2) - φ ∴ K1 = K2 = hc ( (1/λ1) - (1/λ2) ) or, - Δ K = hc ( (1/λ1 ) - (1/λ2) ) or, - 0.52 eV = (1242 eV nm) ((1/500 nm ) - (1/λ2)) or, (-0.52/1242) = (1/500) - (1/λ2) or, (1/λ2) = (1/500) + (0.52/1242) or, λ2 = 413 nm = 400 nm

Q. The kinetic energy of the photoelectrons increases by 0.52eV when the wavelength of incident light is changed from 500nm to another wavelength which is approximately

700 nm

400 nm

1250 nm

1000 nm

Q. The kinetic energy of the photoelectrons increases by $0.52 e V$ when the wavelength of incident light is changed from $500 nm$ to another wavelength which is approximately