Question

A beam of light has three wavelengths $4144\stackrel{\circ }{A},4972\stackrel{\circ }{A}$ and $6216\stackrel{\circ }{A}$ with a total intensity of $3.6\times {10}^{-3}W/m^2$ equally distributed amongst the three wavelengths. The beam falls normally on an area $1.0c{m}^2$ of a clean metallic surface of work function $2.3eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds.

• A. $11\times {10}^{11}$.
• B. $12\times {10}^{11}$.
• C. $13\times {10}^{11}$.
• D. $21\times {10}^{11}$.

Answer 1

The correct option is A $11\times {10}^{11}$.

Threshold wavelength,
${\lambda }_0=\frac{hc}{\varphi }$

${\lambda }_0=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{2.3\times (1.6\times {10}^{-19})}=5.404\times {10}^{-7}$

$=5404\mathring{A}$

Photons with wavelength greater than $5404\mathring{A}$ will not eject electrons

Energy=intensity of each $\lambda \times$ Area of the

$=\frac{3.6\times {10}^{-3}}{3}\times \left(1.0c{m}^2\right)=1.2\times {10}^{-7}W$ $(t=1)$

$E=(1.2\times {10}^{-7})\left(2\right)=2.4\times {10}^{-7}J$ $(t=2)$

No. of photons $n_1$ due to ${\lambda }_{1}4144\mathring{A}$

$n_1=\frac{(2.4\times {10}^{-7})(4144\times {10}^{-10})}{(6.63\times {10}^{-34})(3\times {10}^8)}=0.5\times {10}^{12}$

No. of photons $n_2$ due to ${\lambda }_{2}4972\mathring{A}$

$n_1=\frac{(2.4\times {10}^{-7})(4972\times {10}^{-10})}{(6.63\times {10}^{-34})(3\times {10}^8)}=0.575\times {10}^{12}$

$N=n_1+n_2=(0.5+0.575)\times {10}^{12}$

$=1.075\times {10}^{12}$

$=10.75\times {10}^{11}\approx 11\times {10}^{11}$ (A)