Physics · Atomic Physics

JEE Main 2026 — 28 January, Morning Shift — Question 41

The ratio of de Broglie wavelength of a deutron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2 E$ , is $n: 1$ . The value of $n$ is $\_\_\_\_$ . (Assume mass of proton $=$ mass of neutron) :

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \cdot \mathrm{KE}}}$ $\frac{\lambda_{\mathrm{d}}}{\lambda_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\alpha} \cdot \mathrm{KE}_{\alpha}}{\mathrm{m}_{\mathrm{d}} \cdot \mathrm{KE}_{\mathrm{d}}}}=\sqrt{\frac{4 \mathrm{~m} \cdot 2 \mathrm{E}}{2 \mathrm{~m} \cdot \mathrm{E}}}=2: 1$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter