One milliwatt of light of wavelength 4560 A is incident on a cesium surface. Calculate the photoelectric current liberated assuming a quantum efficiency of 0.5 %. Given Planck's constant `h=6.62xx10^(-34)J-s` and velocity of light `c=3xx10^(8)ms^(-1)`.

Options

A. `1.836 xx10^(-6)A`
B. `1.836 xx10^(-7)A`
C. `1.836 xx10^(-5)A`
D. `1.836 xx10^(-4)A`

Answer

A

Explanation

Energy of photon , `=(hc)/lamda`
Power of lamp = P
No. of photons emitted per second , `N = P / (hc//lamda)`
`:.` No. of photons emitted per second , `=(0.5)/(100) N = 1/(200) xx (Plamda)/(hc)`
`:.` Photoelectric current `=1/200 (Plamda)/(hc) xx e`
`=1/200xx((10^(-3))xx(4560xx10^(-10))xx(1.6 xx 10^(-19)))/((6.62 xx10^(-34))xx(3xx 10^8))= 1.836 xx10^(-6)A`