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A radiation of 5000 Å is incident on a m...

A radiation of 5000 Å is incident on a metal surface whose work function is 1.2 eV. Find out the value of stopping potential.
Given `h=6.6 xx 10^(-34) Js,c= 3 xx 10^(6) ms^(-1), e=1.6 xx 10^(19)C`

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`lambda = 5000 Å = 5000 xx 10^(-10) = 5 xx 10^(-7) m`
`phi_(0) = 1.2 eV`
`"Stopping potential" = (V_(0)) = ?`
`h = 6.6 xx 10^(-34) Js`
`c =3 xx 10^8 ms^-1`
`e = 1.6 xx 10^-19 C`
Energy of incident radiation `=(E)=(hc)/lambda`
`E=(6.6 xx 10^(-34) xx3 xx 10^8)/(5xx10^-7)`
`=19.8/5 xx 10^(-34+8+7)`
`E=3.96 xx 10^-19J`
`implies" "E=(3.96 xx 10^-19)/(1.6 xx 10^-19)eV`
`E=2.475 eV`
We knows`" "eV_(0)=E-phi_(0)`
`eV_(0)=2.475-1.2`
`eV_(0)=1.275 eV`
`V_(0)=1.2 "volt"`.
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