Q 12-11-043JEE MainJEE Main 2026 (28 Jan, Shift 1)Easy
The ratio of de Broglie wavelength of a deutron 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 value type. Enter your answer.
Answer: 2
$\lambda = \dfrac{h}{\sqrt{2mK}}$. Taking the nucleon mass as $m$: deuteron mass $2m$, alpha mass $4m$.
$$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{m_\alpha K_\alpha}{m_dK_d}} = \sqrt{\frac{4m\times2E}{2m\times E}} = \sqrt4 = 2$$
So $n = 2$.
Solution by Sreeraj P, M.Sc Physics