Question: Q. 6. The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of a photon is $(2 \lambda m c / h)$ times the kinetic energy of electron; where $m, c$ and $h$ have their usual meaning. [Foreign I, II, III 2016]
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Solution:
Ans. Energy of photon $E=h v=\frac{h c}{\lambda}$
$$ \Rightarrow \quad \frac{h}{\lambda}=\frac{E}{c} $$
de-Broglie wavelength of electron
$$ \lambda=\frac{h}{p} $$
Kinetic energy of electron,
$$ K=\frac{p^{2}}{2 m} $$
$$ \begin{align*} & =\frac{h^{2}}{2 m \lambda^{2}}=\left(\frac{h}{2 m \lambda}\right)\left(\frac{h}{\lambda}\right) \ & =\left(\frac{h}{2 m \lambda}\right)\left(\frac{E}{c}\right) \ \Rightarrow \quad E & =\left(\frac{2 m c \lambda}{h}\right) K \end{align*} $$
[CBSE Marking Scheme 2016]
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