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JEE Main 202628 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Morning Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2026 · 28 January · Morning Shift · Physics · Dual Nature of Matter

The ratio of de Broglie wavelength of a deutron with kinetic energy...

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) :

Correct answer

0

Step-by-step solution

The de Broglie wavelength is

\lambda = \frac{h}{\sqrt{2mE}}

.
For a deuteron (mass

m_d = 2m_p

) with energy

E

:

\lambda_d = \frac{h}{\sqrt{2 \cdot 2m_p \cdot E}} = \frac{h}{\sqrt{4m_p E}}

For an alpha particle (mass

m_\alpha = 4m_p

) with energy

2E

:

\lambda_\alpha = \frac{h}{\sqrt{2 \cdot 4m_p \cdot 2E}} = \frac{h}{\sqrt{16m_p E}}

The ratio:

\frac{\lambda_d}{\lambda_\alpha} = \frac{\sqrt{16m_p E}}{\sqrt{4m_p E}} = \sqrt{4} = 2

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