Contact us
Back

Report Issue

JEE MAIN 2020
06.09.20_S1
Question
An electron, a doubly ionized helium ion ( $\mathrm{He}^{++}$) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_{\mathrm{e}}, \lambda_{\mathrm{e}++}$ and $\lambda_{\mathrm{p}}$ is
Select the correct option:
A
$\lambda_e>\lambda \mathrm{H}_{\mathrm{e}++}>\lambda_{\mathrm{p}}$
B
$\lambda_{\mathrm{e}}<\lambda_{\mathrm{p}}<\lambda_{\mathrm{He}++}$
C
$\lambda_e>\lambda_p>\lambda_{\mathrm{He}++}$
D
$\lambda_{\mathrm{e}}<\lambda_{\mathrm{He}++}=\lambda_{\mathrm{p}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \mathrm{p}=\sqrt{2 \mathrm{mk}} \\ & \lambda \propto \frac{1}{\sqrt{\mathrm{~m}}}\end{aligned}$
Question Tags
JEE Main
Physics
Easy
Start Preparing for JEE with Competishun
YouTube
Books
App
Tests