Question #1fdc3

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Question #1fdc3

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Answer 1 · 2016-10-18T04:20:37

Given

$λ \to The wave length of uv light = 120 n m = 120 \times 10^{- 9} m$ $lambda->"The wave length of uv light"=120nm=120xx10^-9m$

$I \to Intensity of uv light = 40 mW/ c m^{2}$ $I->"Intensity of uv light"=40"mW/"cm^2$
$= 40 \times 10^{- 3} \times 10^{4} W/ m^{2} = 400 W/ m^{2}$ $=40xx10^-3xx10^4"W/"m^2=400"W/"m^2$

$c \to Velocity of light = 3 \times 10^{8} m/s$ $c->"Velocity of light"=3xx10^8"m/s"$

$e \to Charge of an electron = 1.6 \times 10^{- 19} C$ $e->"Charge of an electron"=1. 6xx10^-19C$

$h \to Planck's constant = 6.625 \times 10^{- 34} J s$ $h->"Planck's constant"=6.625xx10^-34Js$

By Plack's theory the energy of a photon,

$E = h ν = \frac{h c}{λ}$ $E=hnu=(hc)/lambda$

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 $m^{2}$ $m^2$ is

$n = \frac{I}{h ν} = \frac{I λ}{h c}$ $n=I/(hnu)=(Ilambda)/(hc)$

$Current density, J = n \times e = \frac{I λ e}{h c}$ $"Current density, "J=nxxe=(Ilambdae)/(hc)$

$= \frac{400 \times 120 \times 10^{9} \times 1.6 \times 10^{- 19}}{6.625 \times 10^{- 34} \times 3 \times 10^{8}} A m^{- 2}$ $=(400xx120xx10^9xx1.6xx10^-19)/(6.625xx10^(-34)xx3xx10^8)Am^-2$

$= 3.86 \times 10^{19} A m^{- 2}$ $=3.86xx10^19Am^-2$

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