Question

The threshold frequency for a certain metal for photoelectric effect is 1.7 x 10¹⁵ Hz. When a light of frequency 2.2 x 10¹⁵ Hz is incident on the metal surface, the kinetic energy of the emitted photoelectrons is 3.3 x.10^-19 J. Calculate Planck's constant.

Answer 1

Given: v₀ = 1.7 x 10¹⁵ Hz, v = 2.2 x 10¹⁵ Hz K.E._max = 3.3 x l0^-19 J,

`1/2mV_(max)^2 = h(v-v_0)`

`h = (1/2mV_(max)^2)/(v-v_0) = (K.E._(max))/(v-v_0)`

`=(3.3xx10^-19)/(2.2xx10^15-1.7xx10^15)`

`h = 6.6xx10-34  Js`