1.

Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of 3V. The photoelectric effect begins in this metal at frequency 6xx10^(14)s^(-1)

Answer»

`10^(5)`
`10^(16)`
`1.324xx10^(15)`
none of these.

Solution :`hv=E_(k)+w "" :. E_(k)=hv-w`
If `V_(s)` is retarding POTENTIAL then
`E_(k)=eV_(s) and w=hv_(0)`
`:.eV_(s)=hv-hv_(0)`
or `v=(eV_(s))/(H)+v_(0)`
Given `V_(s)=3V, v_(0)=6xx10^(-14)s^(-1)`
`:.` REQUIRED frequencyv is
`v=(1*6xx10^(-19)xx3)/(6*62xx10^(-34))+6xx10^(14)`
`v=1*324xx10^(15)` per sec.


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