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JEE MAIN 2021
16-03-2021 S2
Question
The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V . What should nearly be the ratio of their wavelengths? $\left(\mathrm{m}_{\mathrm{P}}=1.00727 \mathrm{u}, \mathrm{m}_{\mathrm{e}}=0.00055 \mathrm{u}\right)$
Select the correct option:
A
1860 : 1
B
$(1860)^2: 1$
C
41.4 : 1
D
43 : 1
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}} \\ & \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{m}_2}{\mathrm{~m}_1}} \\ & \frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}}=\sqrt{1831.4}=42.79\end{aligned}$
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