A light of frequency 1.6 10¹⁶ ~Hz when falls on a metal plate emits electrons that have double the kinetic energy compared to the kinetic energy of emitted electrons when frequency of 1.0 10¹⁶ ~Hz falls on the same plate. The threshold frequency (v₀ ) of the metal in Hz is

Step-by-step solution

When 1.6 10¹⁶ ~Hz frequency falls on metal plate than double kinetic energy obtained. h (1.6 10¹⁶-v₀ )=2 ~K . E Eq. (i) When 1.0 10¹⁶ ~Hz frequency falls on same plate than K.E. h (1.0 10¹⁶-v₀ )= K.E ( Eq. (ii) ) From Eqs. (i) and (ii) aligned & h (1.6 10¹⁶-v₀ ) h (1.0 10¹⁶-v₀ ) = 2 ~K . E K . E & 1.6 10¹⁶-v₀=2 1.0 10¹⁶-v₀ & 1.6 10¹⁶-2 (1.0 10¹⁶ )=4 10¹⁵ ~Hz & v₀=4 10¹⁵ ~Hz aligned