A beam of light has three wavelengths `4144 Å`, `4972 Å` and `6216 Å` with a total intensity 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

We know that threshold wavelength `(lamda_0)=(hc)/(phi)`.
`lamda_0=((6.63xx10^(-34))(3xx10^(8)))/(2.3xx(1.6xx10^(-19)))=5.404xx10^(-7)m` `=5404 A`
Thus wavelengths `4144 A` and `4972 A` will emits electrons from the metal surface.
Energy incident on the surface per unit time for each wavelength
`="intensity of each wavelength" xx "Area of the surface"`
`=(3.6xx10^(-3))/(3)xx(1.0cm^2)=1.2xx10^(-7)W`
Energy incident on the surface for each wavelength in 2 s,
`E=(1.2xx10^(-7))xx(2)=2.4xx10^(-7)J`
Number of photons `n_1` due to wavelength `4144 A (=4144xx10^(-10))m`),
`n_1=((2.4xx10^(-7))(4144xx10^(-10)))/((6.63xx10^(-34))(3xx10^(8)))=0.5xx10^(12)`
Number of photons `n_2` due to wavelength `4972 A`,
`n_2=((2.4xx10^(-7))(4972xx10^(10)))/((6.63xx10^(-34))(3xx10^(8)))=0.575xx10^(12)`
`N=n_1+n_2=0.5xx10^(12)+0.575xx10^(12)`
`=1.075xx10^(12)`