Question

Electromagnetic radiation having $\lambda =310\dot{A}$ is subjected to a metal sheet having work function = 12.8 eV. What will be the velocity of photo electrons having maximum kinetic energy?

• A. $1.23\times {10}^{6}m/s$
• B. $2.28\times {10}^{6}m/s$
• C. $3.1\times {10}^{6}m/s$
• D. None of these

Answer 1

The correct option is C $3.1\times {10}^{6}m/s$

Energy of photon of wavelength, $\lambda =310\times {10}^{-10}$ is $\frac{hc}{\lambda }$.

Therefore ,$E\left(photon\right)=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{(31\times {10}^{-9})}=6.406\times {10}^{-18}J.$

Work function, $W=12.8eV=12.6\times 1.6x{10}^{-19}J=2.016\times {10}^{-18}J.$

Now, maximum kinetic energy of photo electron,

$\left(0.5\right)m{v}_{\left(max\right)}^2=E\left(photon\right)-W=6.406\times {10}^{-18}-2.016\times {10}^{-18}J=4.39\times {10}^{-18}J.$

Therefore,$v_{\left(max\right)}^2=\frac{(4.39\times {10}^{-18})\left(2\right)}{(9.1\times {10}^{-31})}=9.648\times {10}^{12}{m}^2{s}^{-2}.$

Therefore, $v_{\left(max\right)}=3.106\times {10}^{6}m{s}^{-1}$

Hence,option C is correct answer.