Search any question & find its solution
Question:
Answered & Verified by Expert
If the kinetic energy of a particle is reduced to half, de-Broglie wavelength becomes
Options:
Solution:
2715 Upvotes
Verified Answer
The correct answer is:
$\sqrt 2$ times
$\begin{aligned} & \text { } \lambda \propto \frac{1}{\sqrt{\mathrm{K} . E}} \Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{(\mathrm{KE})_2}{(\mathrm{KE})_1}} \\ & \mathrm{KE}_2=\frac{(\mathrm{KE})_1}{2} \\ & \text { Thus, } \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{1}{2}} \Rightarrow \lambda_2=\sqrt{2} \lambda_1\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.