The ratio of de Broglie wavelength of a deuteron with kinetic energy… | JEE Main 2026 (28 Jan, Shift 1)

Q13JEE Main 2026 · 28 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

The ratio of de Broglie wavelength of a deuteron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2E$ , is $n : 1$ . The value of $n$ is ______.

(Assume mass of proton $=$ mass of neutron)

Numerical answer type. Work it out, then open the answer.

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Answer: 2

$\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mK}}$ .
Deuteron: $m = 2m_p$ , $K = E$ , so $mK = 2m_pE$ . Alpha: $m = 4m_p$ , $K = 2E$ , so $mK = 8m_pE$ .
$\dfrac{\lambda_d}{\lambda_\alpha} = \sqrt{\dfrac{8}{2}} = 2$
So $n = 2$ .

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