The light photons of energy 1 eV and 3 eV respectively are made incident on a metallic plate of work function 0.5 eV one after the other. Find the ratio of maximum kinetic energy of photoelectrons emitted by them:

Question

The light photons of energy 1 eV and 3 eV respectively are made incident on a metallic plate of work function 0.5 eV one after the other. Find the ratio of maximum kinetic energy of photoelectrons emitted by them: (A) 1: 10 (B) 2: 15 (C) 1: 5 (D) 4: 5

Tardigrade Admin · Accepted Answer

( E k 1/ E k 2)=( hv 1-φ/ hv 2-φ2) =(1-0.5/3-0.5) (0.5/2.5)=(1/5)

Q. The light photons of energy $1 e V$ and $3 e V$ respectively are made incident on a metallic plate of work function $0.5 e V$ one after the other. Find the ratio of maximum kinetic energy of photoelectrons emitted by them:

1: 10

2: 15

1: 5

4: 5

$\frac{E _{k_{1}}}{E _{k_{2}}} = \frac{h v _{1} - ϕ}{h v _{2} - ϕ _{2}}$
$= \frac{1 - 0.5}{3 - 0.5}$
$\frac{0.5}{2.5} = \frac{1}{5}$

Q. The light photons of energy 1eV and 3eV respectively are made incident on a metallic plate of work function 0.5eV one after the other. Find the ratio of maximum kinetic energy of photoelectrons emitted by them: