Question

A beam of light has two wavelengths $4972\mathring{A}$ and $6216\mathring{A}$ with a total intensity of $3.6\times 10^{-3}W{m}^{-2}$ equally distributed among the two wavelengths. The beam falls normally on an area of $1c{m}^2$ of a clean metallic surface of work function $2.3eV$. 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 $I=3.6\times 10^{-3}W{m}^{-2}$
Intensity associated with each wavelength
$=\frac{3.6\times 10^{-3}}{2}=1.8\times 10^{-3}W{m}^{-2}$
Work function $\varphi =hv=\frac{hc}{\lambda }$
$=\frac{\left(6.62\times 10^{-34}\right)\left(3\times 10^8\right)}{\lambda }$
$=\frac{12.4\times 10^3}{\lambda }ev$
for different wavelengths
${\varphi }_1=\frac{12.4\times 10^3}{{\lambda }_1}=\frac{12.4\times 10^3}{4972}=2.493eV=3.984\ell 10^{-19}J$
${\varphi }_2=\frac{12.4\times 10^3}{{\lambda }_1}=\frac{12.4\times 10^3}{6216}=1.994eV=3.184\times 10^{-19}J$
Work function for metallic surface $\varphi =2.3eV$ (given)
${\varphi }_2<\varphi$
Therefore, ${\varphi }_2$ 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 $m^2-s$
$=\frac{1.8\times 10^{-3}}{3.984\times 10^{-19}}\times 10^{-4}$
$\left(\because 1c{m}^2=10^{-4}{m}^2\right)=0.45\times 10^{12}$
So, the number of photo electrons liberated in $2$ sec.
$=0.45\times 10^{12}\times 2$
$=9\times 10^{11}$