Question

The 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 the area $1c{m}^2$ of a clean metallic surface of work function $2.3eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2s$.

• A. $2\times {10}^9$
• B. $9\times {10}^8$
• C. $1.075\times {10}^{12}$
• D. $3.75\times {10}^6$

Answer 1

Answer: (C)

As we know, threshold wavelength $\left({\lambda }_0\right)=\frac{hc}{\varphi }$
$\Rightarrow {\lambda }_0=\frac{(6.63\times {10}^{-34})\times 3\times {10}^8}{2.3\times (1.6\times {10}^{-19})}=5.404\times {10}^{-7}m$.
$\Rightarrow {\lambda }_0=5404\stackrel{\circ }{A}$
Hence, wavelength $4144\stackrel{\circ }{A}$ and $4972\stackrel{\circ }{A}$ will emit electron from the metal surface.
For each wavelength energy incident on the surface per unit time
$=$ intensity of each $\times$ area of the surface wavelength
$=\frac{3.6\times {10}^{-3}}{3}\times (1cm{)}^2=1.2\times {10}^{-7}joule$
Therefore, energy incident on the surface for each wavelengths is $2s$
$E=(1.2\times {10}^{-7})\times 2=2.4\times {10}^{-7}J$
Number of photons $n_1$ due to wavelength $4144\stackrel{\circ }{A}$
$n_1=\frac{(2.4\times {10}^{-7})(4144\times {10}^{-10})}{(6.63\times {10}^{-34})(3\times {10}^8)}=0.5\times {10}^{12}$
Number of photon $n_2$ due to the wavelength $4972\stackrel{\circ }{A}$
$n_2=\frac{(2.4\times {10}^{-7})(4972\times {10}^{-10})}{(6.63\times {10}^{-34})(3\times {10}^8)}=0.572\times {10}^{12}$
Therefore total number of photoelectrons liberated in $2s$,
$N=n_1+n_2$
$=0.5\times {10}^{12}+0.575\times {10}^{12}$
$=1.075\times {10}^{12}$.