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Light of wavelength 5000 Å and intensity...

Light of wavelength 5000 Å and intensity of `3.96 xx 10^(-3) watt//cm^(2)` is incident on the surface of photo metal . If 1% of the incident photons only emit photo electrons , then the number of electrons emitted per unit area per second from the surface will be

A

`10^(12)` and 5eV

B

`10^(11)` and 5eV

C

`10^(10)` and 5eV

D

`10^(14)` and 10eV

Text Solution

Verified by Experts

The correct Answer is:
B
`(K.E)_("max")=hv-phi_(0)=(10-5)eV=5eV`
No. of photons emitted per second = total energy incident per sec / energy of one photon
`rArr n= (I xx 4)/(hv)=(1xx10^(-4)xx16xx10^(-3))/(10xx1.6xx10^(-19))=10^(12)`
No. of photon ejected `=n xx 10% = 10^(12)xx(10)/(100) = 10^(11)`
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