# Question 1fdc3

Oct 18, 2016

Given

$\lambda \to \text{The wave length of uv light} = 120 n m = 120 \times {10}^{-} 9 m$

$I \to \text{Intensity of uv light"=40"mW/} c {m}^{2}$
$= 40 \times {10}^{-} 3 \times {10}^{4} \text{W/"m^2=400"W/} {m}^{2}$

$c \to \text{Velocity of light"=3xx10^8"m/s}$

e->"Charge of an electron"=1. 6xx10^-19C#

$h \to \text{Planck's constant} = 6.625 \times {10}^{-} 34 J s$

By Plack's theory the energy of a photon,

$E = h \nu = \frac{h c}{\lambda}$

By Einstein's photoelectric effect number of electrons emitted due to incident of light
on metal equal to the no. of photons in the incident raduation.

So number of electrons emitted per sec per ${m}^{2}$ is

$n = \frac{I}{h \nu} = \frac{I \lambda}{h c}$

$\text{Current density, } J = n \times e = \frac{I \lambda e}{h c}$

$= \frac{400 \times 120 \times {10}^{9} \times 1.6 \times {10}^{-} 19}{6.625 \times {10}^{- 34} \times 3 \times {10}^{8}} A {m}^{-} 2$

$= 3.86 \times {10}^{19} A {m}^{-} 2$