Question

The wavelength of a photon is $1.4A^{\circ }$. It collides with an electron. Its wavelength after the collision is $2.0A^{\circ }$. Calculate the energy of the scattered electron.
Report n if the answer in joule is $n\times {10}^{-16}$
Here plank's constant $h=6.63\times {10}^{-34}{\text{m}}^2\text{kg/s}$

Answer 1

We are given that, initial wavelength of the photon,
$\lambda =1.4A^{\circ }=1.4\times {10}^{-10}m$
Final wavelength of the photon, ${\lambda }^{\prime }=2.0A^{\circ }=2\times {10}^{-10}m$
Initial energy of the photon = $hv=\frac{hc}{\lambda }$
Final energy of the photon $=h{v}^{\prime }=\frac{hc}{{\lambda }^{\prime }}$
If E is the energy of the scattered electron,
$E=\frac{hc}{\lambda }-\frac{hc}{{\lambda }^{\prime }}=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}]=hc\left[\frac{{\lambda }^{\prime }-\lambda }{\lambda {\lambda }^{\prime }}\right]$
$=(6.63\times {10}^{-34})(3\times {10}^8)\left[\frac{2\times {10}^{-10}-1.4\times {10}^{-10}}{(2\times {10}^{-10})(1.4\times {10}^{-10})}\right]$
$=(6.63\times {10}^{-34})(3\times {10}^8)\left(\frac{0.6\times {10}^{-10}}{2.8\times {10}^{-20}}\right)=4.26\times {10}^{-16}J$

Why this question? Tip: i)The energy lost by photon is gained by electron. ii) Some straight forward simple questions are asked even in JEE Advanced. This was not to miss out!