Question

A beam of light has three $\lambda$, $4144\mathring{A},4972\mathring{A}$ and $6216\mathring{A}$ with a total intensity of $3.6\times {10}^{-3}$ ${Wm}^{-2}$ equally distributed amongst the three $\lambda$. The beam falls normally on an area $1.0{cm}^2$ of a clean metallic surface of work function $2.3eV$. Assume that there is no loss of light by reflection etc. Calculate the no. of photoelectrons emitted in $2sec$.

Answer 1

$E_1$ = 12400/4144 = 2.99 ev

$E_2$ = 12400/4972 = 2.49 ev

$E_3$ = 12400/6216 = 1.99 ev

So only first two wavelength are capable of ejecting photoelectrons

Energy incident per second = $3.6/3\times {10}^{-3}\times {10}^{-4}=1.2\times {10}^{-7}$ J/s

$n_1=1.2\times {10}^{-7}/2.99\times 1.6\times {10}^{-19}=2.5\times {10}^{11}$

$n_2=1.2\times {10}^{-7}/2.49\times 1.6\times {10}^{-19}=3\times {10}^{11}$

Total no of photons$=2(n_1+n_2)$

$=3.01\times {10}^{11}+2.51\times {10}^{11}$

So total no photoelectrons ejected in two seconds$=11\times {10}^{11}$