An electron, a doubly ionized helium ion (He++) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths \lambda e, \lambda He++ and \lambda p is :

\lambda e > \lambda p > \lambda He++

\lambda e > \lambda He++ > \lambda p

\lambda e < \lambda p < \lambda He++

\lambda e < \lambda He++ = \lambda p

Dual Nature of Radiation PYQ — JEE Main 2020 · 6 Sep · Shift 1 · Q57 with Solution

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Correct answer: C

For a particle,

\lambda = \frac{h}{p}

where $p$ is momentum.

Since all three particles have the same kinetic energy $K$ , and for non-relativistic motion,

K = \frac{p^2}{2m}

So,

p = \sqrt{2mK}

Hence,

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

Since $h$ and $K$ are same for all particles,

\lambda \propto \frac{1}{\sqrt{m}}

So, smaller mass means larger de-Broglie wavelength.

Electron mass $m_e$ is the smallest.
Proton mass $m_p$ is much larger than electron mass.
Doubly ionized helium ion $\mathrm{He}^{++}$ has mass approximately $4m_p$ (very close, neglecting small electron mass difference).

Thus,

m_e < m_p < m_{\mathrm{He}^{++}}

Therefore,

\lambda_e > \lambda_p > \lambda_{\mathrm{He}^{++}}

This corresponds to:

Option C:

\lambda_e > \lambda_p > \lambda_{\mathrm{He}^{++}}

Stored correct answer: C

This matches our derived answer.