Q.9:- A photon of wavelength 4×10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate
(i) the energy of the photon (eV),
(ii) the kinetic energy of the emission, and
(iii) the velocity of the photoelectron (1 eV = 1.6020×10^-19 J).

Answer:-

(i) Energy of the photon (E) = hν = hc/λ = (6.626×10^-34 Js×3.0×10⁸ ms^-1)/4×10^-7 m = 4.97×10^-19 J
= 4.97×10^-19/1.602×10^-19 eV
(ii) Kinetic energy of emission (1/2 mv²) = hν- hνo = 3.10-2.13 = 0.97 eV
(iii) 1/2 mv² = 0.97 eV = 0.97×1.602×10^-19 J
⇒ 1/2×(9.11×10^{-31 kg})×v² = 0.97×1.602×10^-19 J
⇒ v² = 0.341×10¹² = 34.1×10¹⁰
⇒ v = 5.84×10⁵ ms^-1