Question: 26: The threshold frequency of a photoelectric metal is v_{0}. If light of frequency 4 v_{0} is incident on this metal, then the maximum kinetic energy of emitted electrons will be :
(1) h v_{0}
(2) 2 h v_{0}
(3) 3 h v_{0}
(4) 4 h v_{0}
Answer: Option (3)
Explanation:
According to Einstein’s photoelectric equation, the maximum kinetic energy of emitted electrons is given by
K_{\max} = h\nu - h\nu_{0}Here, the frequency of incident light is
\nu = 4\nu_{0}Substituting this value in the photoelectric equation,
K_{\max} = h(4\nu_{0}) - h\nu_{0} K_{\max} = 3h\nu_{0}Therefore, the maximum kinetic energy of the emitted electrons is 3 h v_{0},
and the correct option is (3).