Search any question & find its solution
Question:
Answered & Verified by Expert
A proton and an electron have the same de Broglie wavelength. If $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{e}}$ be the kinetic energies of proton and electron respectively, then choose the correct relation :
Options:
Solution:
2936 Upvotes
Verified Answer
The correct answer is:
$\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{e}}$
De Broglie wavelength of proton \& electron $=\lambda$
$\begin{aligned} & \because \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \therefore \mathrm{p}_{\text {proton }}=\mathrm{p}_{\text {electron }} \\ & \because \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \\ & \therefore \mathrm{KE}_{\text {proton }} < \mathrm{KE}_{\text {electron }} \\ & {\left[\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{c}}\right]}\end{aligned}$
$\begin{aligned} & \because \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \therefore \mathrm{p}_{\text {proton }}=\mathrm{p}_{\text {electron }} \\ & \because \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \\ & \therefore \mathrm{KE}_{\text {proton }} < \mathrm{KE}_{\text {electron }} \\ & {\left[\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{c}}\right]}\end{aligned}$
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.