Question

A beam of light has two wavelengths $4972A^o$ and $6216A^o$ 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)

$9\times {10}^{11}$

The threshold wavelength is given as $\lambda =\frac{hc}{\varphi }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2.3\times 1.6\times {10}^{-19}}=5380A^o$

Thus the beam with wavelength $6216A^o$ would not emit any electrons from the metal.

Now the intensity of each beam is

$I=\frac{1}{2}\times 3.6\times {10}^{-3}W{m}^{-2}=1.8\times {10}^{-3}W{m}^{-2}$

$Now,forwavelength4972A^o$

$1.8\times {10}^{-3}=\frac{nhC}{\lambda }$

Or,

$n=4.5\times {10}^{15}afterputtingthevalues\left(i\right)$

Now number of photoelectrons emitted from the area $A={10}^{-4}{m}^2$ in time $t=2sec$ is given as

$n_e=nAt=4.5\times {10}^{15}\times {10}^{-4}\times 2=9\times {10}^{11}electrons$

Answer is B.