Answer to Question #236468 in Quantum Mechanics for Zeedan

Question

A ray of light of wavelength 36.5x10^-8 m and intensity of 10^-8 Wm^-2 is incident perpendicularly on a surface. Absorption coefficient of that surface is 0.8 and work function is 1.6eV. Calculate

(a) rate of production of electron from unit area

(b) absorbed energy per square meter

(c) kinetic energy of the produced electrons

Answer 1

Given:

$\lambda=36.5*10^{-8}\:\rm m$

$I=10^{-8}\:\rm W/m^2$

$k=0.8$

$\phi=1.6\:\rm eV=2.56*10^{-19}\:\rm J$

(a) the rate of production of electron from unit area is given by

$$\frac{dN}{Adt}=\frac{kI}{E_0}=\frac{kI\lambda}{hc}$$

$$\frac{dN}{Adt}=\frac{0.8*10^{-8}*36.5*10^{-8}}{6.63*10^{-34}*3.0*10^{8}}=1.47*10^{10} \:\rm 1/m^2\cdot s$$

(b) the absorbed energy per square meter is given by

$$E/At=kI=0.8*10^{-8}=8.0*10^{-9}\:\rm W/m^2$$

(c) the kinetic energy of the produced electrons is given by

$$E_k=hc/\lambda-\phi$$

$$E_k=\frac{6.63*10^{-34}*3.0*10^{8}}{36.5*10^{-8}}-2.56*10^{-19}\\ =2.88*10^{-19} \:\rm J=1.8\:\rm eV$$