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The magnetic field associated with a lig...

The magnetic field associated with a light wave is given, at the origin, by
`B=B_(0)[sin(3.14xx10^(7))ct +sin(6.28xx10^(7))ct].`
If this light falls on a silver plate having a work function fo 4.7 eV, what will be the maximum kinetic energy of the photo electrons?
`(c=3xx10^(8)ms^(-1),h=6.6xx10^(-34)J-s)`

Text Solution

Verified by Experts

According to question there are two EM waves with different frequency
`B_(1)=B_(2)sin (pixx10^(7)c)t`
and `B_(2)=B_(0)sin (2 pi xx10^(7)c)t`
To get maximum kinetic energy we take the photon with higher frequency
using `B=B_(0) sin omega t ` and `omega=2pi vimpliesv=(omega)/(2pi)`
`B_(1)=B_(0)sin (pixx10^(7)c)timpliesv_(1)=(10^(7))/2xxc`
`B_(2)=B_(0)sin(2pixx10^(7)c)timpliesv_(2)=10^(7)c`
where c is speed of light `c=3xx10^(8)m//s`
Clearly `v_(2)gtv_(1)`
so KE of photoelectron will be maximum for photon of higher energy.
`v_(2)=10^(7)cHz`
`hv=phi+KE_("max")`
energy of photon
`E_(ph)=hv=6.6xx10^(-34)xx10^(7)xx3xx10^(9)`
`E_(ph)=6.6xx3xx10^(-19)J`
`=(6.6xx3xx10^(-19))/(1.6xx10^(-19))=3V=12.375eV`
`KE_("max")=E_(ph)-phi`
`=12.375-4.7=7.675eV~~7.7eV`
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