What is the exact ratio of the de Broglie wavelengths of a proton and an -particle if both are accel — Atomic Structure Chemistry Question

💡 Solution & Explanation

The de Broglie wavelength of a particle accelerated by potential $V$ is $\lambda = \frac{h}{\sqrt{2mqV}}$. Therefore, $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$. Since $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$, the ratio becomes $\sqrt{\frac{4m_p \cdot 2e}{m_p \cdot e}} = \sqrt{8} = 2\sqrt{2}$. The ratio is $2\sqrt{2} : 1$.