Question

A beam of light has two wavelengths $4972 \,\mathring{A}$ and $6216 \,\mathring{A}$ with a total intensity of $3.6\times 10^{-3}\, Wm^{-2}$ equally distributed among the two wavelengths. The beam falls normally on an area of $1 \,cm^2$ of a clean metallic surface of work function $2.3\, eV$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in $2s$ is approximately :

• A. $6 \times 10^{11}$
• B. $9 \times 10^{11}$
• C. $11 \times 10^{11}$
• D. $15 \times 10^{11}$

Answer 1

Answer: (B)

Given, $\lambda_{1}=4972\,\mathring{A}$
and $\lambda_{2}=6216\,\mathring{A}$
and $I=3.6\times10^{-3}Wm^{-2}$
Intensity associated with each wavelength
$=\frac{3.6\times10^{-3}}{2}=1.8\times10^{-3}\,Wm^{-2}$
Work function $\phi=hv=\frac{hc}{\lambda}$
$=\frac{\left(6.62\times10^{-34}\right)\left(3\times10^{8}\right)}{\lambda}$
$=\frac{12.4\times10^{3}}{\lambda}ev$
for different wavelengths
$\phi_{1}=\frac{12.4\times10^{3}}{\lambda_{1}}=\frac{12.4\times10^{3}}{4972}=2.493\,eV=3.984\ell10^{-19}\,J$
$\phi_{2}=\frac{12.4\times 10^{3}}{\lambda _{1}}=\frac{12.4\times 10^{3}}{6216}=1.994\,eV=3.184\times10^{-19}\,J$
Work function for metallic surface $\phi=2.3\,eV$ (given)
$\phi_{2}<\phi$
Therefore, $\phi_{2}$ will not contribute in this process.
Now, no. of electrons per $m^{2}-s=$ no. of photons per $m^{2}-s$
no. of electrons per $m^{2}-s$
$=\frac{1.8\times10^{-3}}{3.984\times10^{-19}}\times10^{-4}$
$\left(\because1\,cm^{2}=10^{-4}\,m^{2}\right)=0.45\times10^{12}$
So, the number of photo electrons liberated in $2$ sec.
$=0.45\times10^{12}\times2$
$=9\times10^{11}$