A beam of light has three wavelengths `4144 Å`, `4972 Å` and `6216 Å` with a total instensity of `3.6 xx 10^(-3) Wm^(-2)` equally distributed amongst the three wavelengths. The beam falls normally on an area `1.0 cm^2` of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects on electron. Calculate the number of photo electrons liberated in two seconds.

Explanation

Here, `phi_(0) =2.3eV=2.3xx1.6xx10^(-19)J`
Thereshold wavelength,
`lambda_(0)=(hc)/(phi_0)=((6.63xx10^(-34))xx(3xx10^(8)))/(2.3xx1.6xx10^(-19))`
`=5.404xx10^(-7)=5404Å`
This shows that the beam of light of wavelength `4144Å and 4972Å` will emit photoelectrons from the metal surface.
The enery incident on the surface for each wavelength of light =intensity of each wavelengthx area of surface
`=((3.6xx10^(-3)))/3xx(1.0xx10^(-4))=1.2xx10^(-7)wat t`
Energy incident on the surface for each wavelength of light in 2 second is
`E=(1.2xx10^(-7))xx2=2.4xx10^(-7)joule`
No. of photons of wavelength `lambda` are
`n=E/(hc//lambda)=(Elambda)/(hc)`
No. of photons `n_(1)` due to wavelength `4144Å` is
`n_(1)=((2.4xx10^(-7))xx(4144xx10^(-10)))/((6.63xx10^(- 34))xx(3xx10^(8)))=0.5xx10^(12)`
No. of photons `n_(2)` due to wavelength `4972Å` is
`n_(2)=((2.4xx10^(-7))xx(4972xx10^(-10)))/((6.63xx10^(-34))xx(3xx10^(8)))=0.576xx10^(12)`
Total number of photoelectrons emitted =total number of photons falling which cause photon emission
`=n_(1)+n_(2)=0.5xx10^(12)+0.576xx10(12)`
`=1.075xx10^(12)`