1.

`10^-3` W of `5000A` light is directed on a photoelectric cell. If the current in the cell is `0.16muA`, the percentage of incident photons which produce photoelectrons, isA. 0.004B. 0.0004C. 0.2D. 0.1

Answer» Correct Answer - B
Current, `l=(q)/(t)=(Ne)/(t)=n_(e)e`
where, `n_(e)` = number of electrons per second
`e = 1.6 xx 10^(19)C`
As, each electron is emitted due to absorption of a photon,
`therefore n_(e)=n_(2)` = number of photons absorbed per second
`implies " "n_(r)=(l)/(e)=(0.16xx10^(-6))/(1.6xx10^(-19))"per sec"= 10^(12)" photon"//s`
And power of source,
`P=n_(1)E, n_(1)` = number of photons emitted per second
`implies" "n_(1)=(P)/(E)`
`"where, "E=(hc)/(lambda)=(12400)/(5000)eV`
`impliesE=2.48eV=2.48xx1.6 xx10^(-19)J=3.968 xx10^(-19)J`
`impliesn_(1)=(P)/(E)=2.52xx10^(15)" photons"//s`
`therefore %` of photons absorbed = `(n_(2))/(n_(1))xx100`
`=(10^(12)xx100)/(2.52xx10^(15))=0.03968% ~~0.04%`


Discussion

No Comment Found

Related InterviewSolutions