A light source emitting three wavelength `5000Å, 6000Å, 7000Å` has a total power of `10^(-3)` m. W and the beam of diameter `2xx10^(-3)`m. The power density is distributed equally amongst the three wavelengths. The beam shines normally on a metallic surface of area `10^(4)m^(2)` which has a work function of 1.9 eV. Assuming that each proton liberates an electron, calculate the charge emitted per unit area in one second :

Options

A. 94 C
B. 54 C
C. 47 C
D. 74 C

Answer

A

Explanation

The threshold wavelngth `lambda_(0)` of metallic surface is given by
`w=(hc)/(lambda_(0)) :. Lambda_(0)=6513Å`
For photoelectric emission `lambda lt lambda_(0)`, so cut of the given
wavelength `lambda_(1)=5000Å and lambda_(2)=6000Å` will cause photo emission.
Power of source `=10^(-3)W`
Diameter of beam `D=2xx10^(-3)m`
Radius of beam `r=10^(-3)m`
`:.` Area of cross section of beam
`A=pi r^(2)=pi xx10^(6)m^(2)`
Power density of beam `=(P)/(A)`
`=(10^(-3))/(pixx10^(-6))=(1)/(pi)xx10^(3)W//m^(3)`
Power density of each wavelength,
`I.=(10^(3))/(3pi)W//m^(3)`
Energy of photon of wavelength `lambda_(1)`
`E_(1)=(hc)/(lambda_(1))=3.97xx10^(-19)J`
Number of photons of wavelength `lambda_(1)` incident on metallic surface per unit area per sec.
`N_(1)=(I)/(E_(1))=(10^(3))/(3pi xx3.97xx10^(-19))=2.67xx10^(20)`
Energy of photon of wavelength `lambda_(2)`
`E_(2)=(hc)/(lambda_(2))=3.31xx10^(-19)J`
Then `N_(2)=(I.)/(E_(2))=3.21xx10^(20)`
As each photon liberates an electron, the number of electrons emitted per sec unit area
`N=N_(1)+N_(2)`
`=2.67xx10^(20)+3.21xx10^(20)`
`=5.88xx10^(20)`
`:.` Charge emitted per sec. per unit area is
`q=Ne`
`=5.88xx10^(20)xx1.6xx10^(-19)`
`or q=94C`