1.

The photoelectric surface is receiving light of wave length 5000Å at the rate of10^(-7) J/s. The no. of photoelectron received per sec is:

Answer»

`2.5xx10^(12)`
`2.5xx10^(11)`
`2.5xx10^(10)`
`2.5xx10^(9)`

SOLUTION :photns per sec `(10^(-7))/(hc//LAMBDA)=(10^(-7))XX (lambda)/(hc)`
`=(10^(-7)xx5000xx10^(-10))/(6.6xx10^(-34)xx3xx10^(8))`
`=2.5xx10^(11)`


Discussion

No Comment Found

Related InterviewSolutions