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. Ratio of maximum speeds of emitted electrons will be

Question

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. Ratio of maximum speeds of emitted electrons will be (A) 1: 5 (B) 1: 4 (C) 1: 2 (D) 1: 1

Tardigrade Admin · Accepted Answer

v max =√(2/ m ) K ⋅ E max v max =√(2/ m )( E Ph - W ) ( v 1/ v 2) =√( E Ph 1- W / E Ph 2- W ) =√(1-0.5/2.5-0.5) ( v 1/ v 2) =(1/2)

Q. Light of two different frequencies whose photons have energies $1 e V$ and $2.5 e V$ respectively, successively illuminate a metallic surface whose work function is $0.5 e V$ .
Ratio of maximum speeds of emitted electrons will be

1 : 5

1 : 4

1 : 2

1 : 1

$v_{m a x} = \frac{2}{m} K \cdot E_{m a x}$
$v_{m a x} = \frac{2}{m} (E_{P h} - W)$
$\frac{v _{1}}{v _{2}} = \frac{E _{P h_{1}} - W}{E _{P h_{2}} - W}$
$= \frac{1 - 0.5}{2.5 - 0.5}$
$\frac{v _{1}}{v _{2}} = \frac{1}{2}$

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