Physics - Modern Physics Question with Solution | TestHub

The de Broglie wavelength is given by $\lambda =\frac{h}{\sqrt{2mqV}}$.

Since the de Broglie wavelengths are the same, ${\lambda }_{\alpha }={\lambda }_p$.

$\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }{V}_{\alpha }}}=\frac{h}{\sqrt{2m_p{q}_p{V}_p}}$

$m_{\alpha }{q}_{\alpha }{V}_{\alpha }=m_p{q}_p{V}_p$

$\frac{V_{\alpha }}{V_p}=\frac{m_p{q}_p}{m_{\alpha }{q}_{\alpha }}$

Given $m_{\alpha }=6.4\times 10^{-27}$ kg and $m_p=1.6\times 10^{-27}$ kg, we have $m_{\alpha }=4m_p$.

The charge of an $\alpha$-particle is twice that of a proton, so $q_{\alpha }=2q_p$.

$\frac{V_{\alpha }}{V_p}=\frac{m_p\times q_p}{4m_p\times 2q_p}=\frac{1}{8}$

The ratio is $1:8$.

Answer: $1:8$