A uniform monochromatic beam of light of wavelength `365xx10^(-9)`m and intensity `10^(-8) Wm^(-2)` falls on a surface having avsorption coefficient 0.8 and work functoin1.6 eV. Determine the rate of number of electrons emitted per `m^(2)`, power absorbed per `m^(2)` and the maximum kinetic energy of emitted photoelectrons.

Explanation

The intensity of radiation I, is defined as the energy passing per unit time per unit area normal to the direction of the beam, If N be the number of photons crossing unit area per unit time, the `I=Nxx"energy carried by one photon"`
`I=N(hc)/(lamda)` or `N=(Ilamda)/(hc)`
If `N_i` be the number of incident photons per unit area per unit time, then
`N_i=(I_ilamda)/(hc)=(10^(-8)xx365xx10^(19))/(6.62xx10^(-34)xx3xx10^(8))=19=18.35xx10^(9)`
The number of photons absorbed, `N_(ab)`, by the surface per unit area per unit time is given by `N_(ab)=absorption coeffiecient of surface xxN_(i)`
`=0.8xx18.35xx10^(8)=1.47xx10^(10)m^(-2)s^(-1)`
Now, assuming that each photon ejects only one electron, the rate of electrons emitted per unit area is given by
`N=N_(ab)=1.47xx10^(10)m^(-2)s^(-1)`
power absorbed per `m^2="Absorption coefficient"xx("incident")/(m^(2))=0.8xx10^(-8)=8xx10^(-9)Wm^(-2)`
From Eienstein's equation, maximum kinetic energy is given by `KE_(max)=hv-W_0=(hc)/(lamda)-W_0`
`=((6.62xx10^(-34))(3xx10^(8)))/(365xx10^(-9))-1.6xx1.6xx10^(19)`
`=2.89xx10^(-19)J=1.80 eV`