The threshold frequency of a metal is 10¹⁵ ~s ⁻¹ . The ratio of maximum kinetic energies of the photoelectrons, when the metal is made to strike with radiations of frequencies 1.5 10¹⁵ ~s ⁻¹ and 2.0 10¹⁵ ~s ⁻¹ respectively is

Step-by-step solution

v₀=1.0 10¹⁵ ~s ⁻¹ Maximum kinetic energy will be at the highest frequencies and when all the energy is utilized for the kinetic energy of the ejected electrons: aligned & K . E ₁= h (v₁-v₀ ) & =6.626 10⁻³⁴ (1.5 10¹⁵-1.0 10¹⁵ ) & =6.626 10⁻³⁴ (0.5 10¹⁵ ) & =3.313 10⁻¹⁹ ~J & K _ . E ₂= h (v₂-v₀ ) & =6.626 10⁻³⁴ (2.0 10¹⁵-1.0 10¹⁵ ) & =6.626 10⁻³⁴ (1 10¹⁵ ) & =6.626 10⁻¹⁹ ~J aligned Thus, the ratio will be : K . E ₁ ~K . E ₂ = 3.313 10⁻¹⁹ 6.626 10⁻¹⁹ =1: 2