Light of two different frequencies whose photons have energies 1 eV and 2.5 eV , respectively, successively illuminate a metallic surface whose work function is 0.5 eV . The ratio of maximum speeds of the emitted electrons will be

Step-by-step solution

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is K m a x = h ν - ϕ 0 Where h ν is the energy of the incident photon and ϕ 0 is the work function. But K m a x = 1 2 m v m a x 2 ∴ 1 2 m v m a x 2 = h ν - ϕ 0 As per the question, 1 2 m v 2 max 1 ⁡ = 1 eV - 0.5 eV = 0.5 eV ...(i) and 1 2 m v 2 max 2 ⁡ = 2.5 eV - 0.5 eV = 2 eV ...(ii) Dividing equation (i) by equation (ii), we get v 2 max 1 ⁡ v 2 max 2 ⁡ = 0.5 eV 2 eV = 1 4 v max 1 ⁡ v max 2 ⁡ = 1 4 = 1 2