Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If the kinetic energies of an electron, an alpha particle and a proton having same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$ and $\varepsilon_3$ respectively, then
PhysicsDual Nature of MatterJEE Main
Options:
  • A $\varepsilon_1>\varepsilon_3>\varepsilon_2$
  • B $\varepsilon_1=\varepsilon_2=\varepsilon_2$
  • C $\varepsilon_1 < \varepsilon_3 < \varepsilon_2$
  • D $\varepsilon_1>\varepsilon_2>\varepsilon_3$
Solution:
1427 Upvotes Verified Answer
The correct answer is: $\varepsilon_1>\varepsilon_3>\varepsilon_2$
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}, \sqrt{\mathrm{mk}}=\mathrm{const}$
$\mathrm{k} \propto \frac{1}{\mathrm{~m}}, \varepsilon_1>\varepsilon_3>\varepsilon_2$

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.