MP_JEE MAIN_16-03-2021_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

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:

1860 : 1

$(1860)^2: 1$

41.4 : 1

43 : 1

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}} \\ & \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{m}_2}{\mathrm{~m}_1}} \\ & \frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}}=\sqrt{1831.4}=42.79\end{aligned}$

Question Tags

JEE Main

Physics

Easy

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