De Broglie Wavelengths Of Different Charged Particles

For a charge $q$ accelerated from rest through potential $V$, the kinetic energy is $qV$ and the momentum is $p=\sqrt{2mqV}$, giving a de Broglie wavelength $\lambda =\frac{h}{\sqrt{2mqV}}$. Since $V$ is the same for both particles, $\lambda \propto \frac{1}{\sqrt{mq}}$, so the ratio depends only on mass and charge. The alpha particle has mass $4m_p$ and charge $2e$, while the proton has mass $m_p$ and charge $e$. Therefore $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}}=\sqrt{\frac{4\times 2}{1\times 1}}=\sqrt{8}=2\sqrt{2}$. The option $\sqrt{2}:1$ accounts for charge but ignores the mass factor. The option $1:2\sqrt{2}$ inverts the correct ratio. The option $4:1$ uses mass only and drops the square root. Because the alpha particle is both heavier and more highly charged, it carries far more momentum at the same voltage and hence a shorter wavelength, so the proton's wavelength is the larger one, consistent with our ratio exceeding unity. This combined mass-charge dependence is a favourite JEE extension of the basic accelerated-particle formula. A deeper point is that the combined mass-and-charge factor under the square root makes such comparisons sensitive to both how heavy a particle is and how strongly the field accelerates it. Had the two particles instead been given the same kinetic energy rather than the same accelerating voltage, only the mass ratio would matter, so identifying the correct constraint before substituting is essential to avoid a common error.