A light source, emitting there wavelengths 500nm, 600nm and 700 nm, has a total power of `10^(-3) W` and a beam diameter `2xx10^(-3)m`. The density is distributed equally amongst the three wavelength. The beam shines normally on a metallic surface of area on `10^(-4) m^(2)` and having a work function of `1.9` eV. Assuming that each photon liberates an electron, calculate the charge emitted per unit area in one second.

Explanation

Work function `phi =1.9 eV`
Threshold wavelength `lambda_(0) =(1242)/(phi(eV))nm`
`=(1242)/(1.9) =654nm`
The photoelectric will take place for wavelength 500 nm, 600 nm and 700 nm `(700 gt 654)`
Intensity of incident beam
`I=(P)/(A) =(10^(-3))/(pi(10^(-3))^(2)) =(1000)/(pi)W//m^(2)`
Intensity is distributed equally in three wavelengths `I_(1)=I_(2)=I_(3)=I//3`
The emission place for `lambda_(1)=500 nm, lambda_(2)=600 nm` Number of photons`//m^(2)//`sec for these wavelengths
`n_(1)+n_(2) =(I//3)/(hc//lambda_(1)) + (I//3)/(hc//lambda_(2)) =(I)/(3hc)(lambda_(1)+lambda_(2))`
`=(1000(500+600)xx10^(-9))/(pixx3xx6.6xx10^(-34)xx3xx10^(8))`
`5.9xx10^(20)` = number of electrons `//m^(2)//`sec

Beam area is smaller than surface area
`q=5.9xx10^(20)e`
`=5.9xx10^(20)xx1.6xx10^(-19)`
`=94.4 C//m^(2)//sec`