Work Power Energy Question 4
Question 4 - 2024 (29 Jan Shift 1)
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25 %$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:
(1) $\frac{1}{1}$
(2) $\frac{1}{8}$
(3) $\frac{8}{1}$
(4) $\frac{1}{4}$
Show Answer
Answer: (2)
Solution:
For photon
$\mathrm{E}{\mathrm{p}}=\frac{\mathrm{hc}}{\lambda{\mathrm{p}}} \Rightarrow \lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{p}}}$
For electron
$\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}{\mathrm{e}} \mathrm{v}{\mathrm{e}}}=\frac{\mathrm{hv}{\mathrm{e}}}{2 \mathrm{~K}{\mathrm{e}}}$
Given $\mathrm{v}_{\mathrm{c}}=0.25 \mathrm{c}$
$\lambda_{\mathrm{e}}=\frac{\mathrm{h} \times 0.25 \mathrm{c}}{2 \mathrm{~K}{\mathrm{e}}}=\frac{\mathrm{hc}}{8 \mathrm{~K}{\mathrm{e}}}$
Also $\lambda_{\mathrm{p}}=\lambda_{\mathrm{e}}$
$\frac{\mathrm{hc}}{\mathrm{E}{\mathrm{p}}}=\frac{\mathrm{hc}}{8 \mathrm{~K}{\mathrm{e}}}$
$\frac{\mathrm{K}{\mathrm{e}}}{\mathrm{E}{\mathrm{p}}}=\frac{1}{8}$
