Saved Bookmarks
| 1. |
(a)1 mW of light of wavelength 456 nm is incident on a cesium surface. Calculate the photoelectric current produced, if the quantum efficiency of the surface for photoelectric emission is only 0.5 %. (b) In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches its saturation value. Assuming that on the average one out of every 10^(6) photons is able to eject a photoelectron, find the photocurrent in the circuit. |
|
Answer» Solution :(a) `P =n(hc)/(lambda)`, n: NUMBER of photons/sec `n=(P lambda)/(h c)` Quantum efficiency `eta =("number of electrons ejected"//"sec")/("number of photons incident"//sec") =(n')/(n)` `n' eta n` Photoelectric CURRENT `I =n' E=eta n e` `=eta(P lambda)/(hc)e` `=((0.5)/(100)) ((10^(-3))(456xx10^(-9)))/((6.6xx10^(-34))(3xx10^(8)))xx1.6xx10^(-19)` `=1.84xx10^(-6) A=1.84 mu A` (b) Here `eta=(1)/(10^(6))` `i=eta(P lambda)/(hc)e =(1)/(10^(6))xx(5xx400xx10^(-9))/(6.6xx10^(-34)xx3xx10^(8))xx1.6xx10^(-19)` `=1.6xx10^(-6) A=1.6 mu A` |
|