A beam of light has three wavelengths `4000 Å, 5000 Å, 6000Å` with a total intensity `3 xx 10^(-3) W//m^(2)` equally distributed amongst the three wavelenth. The beam falls normally on an area `2 cm^(2)` of clean metallic surface of work function `2.4 eV`. Calculate photo current. (Assume each energetically suitable photon emits one electron)

Answer

`0.14 muÅ`

Explanation

`(hc)/(lambda_(1)) = (12400)/(4000) eV = 3.1 eV`
`(hc)/(lambda_(2)) = (12400)/(5000) eV = 2.48 eV`
`(hc)/(lambda_(2)) = (12400)/(6000) eV = 2.06 eV`
`:.` Light having wavelength `6000 7` will not be able to eject electrons.
`:.` Photo-current `= (overset(.)(n)_(1) + overset(.)(n)_(2))e`
where `overset(.)(n)_(1)=` no. of photons of light incident having wavelength `4000Å`
`overset(.)(n)_(2)` no. of photons of light incident having wavelength `5000 Å`. Here
`overset(.)(n)_(1) ((hc)/(lambda_(1))) = overset(.)(n)_(2) ((hc)/(lambda_(2))) = overset(.)(n)_(3) ((hc)/(lambda_(3))) = (I_(0))/(3) = (3 xx 10^(-3))/(3) = 10^(3)`
Area intercepted `= 2cm^(2) = 2 xx 10^(-4)m^(2)`
`:.` Photo-current `= (overset(.)(n)_(1) + overset(.)(n)_(2))e`
`= ((10^(-3) xx 2 xx 10^(-4))(5000 + 4000) xx 10^(-10) xx 1.6 xx 10^(-19))/(6.63 xx 10^(-34) xx 3 xx 10^(8))`
`= 0.144 muA`