Recall the formula of de-Broglie wavelength,

$\lambda =\frac{h}{p}$, where $h \& p$ are Planck's constant and momentum of a particle, here,

$m_{\alpha }=4m_p \& p=\sqrt{2mqV}$ where $V$ is the potential applied and $q$ is charged in particle, 

so the ration fo de-Broglie wavelength,

$\frac{{\lambda }_{\alpha }}{{\lambda }_p}=\frac{{\displaystyle \frac{h}{\sqrt{2m_{\alpha }qV}}}}{\frac{h}{\sqrt{2m_{p}qV}}}=\frac{{\displaystyle \frac{h}{\sqrt{2\times 4\times m_p\times 2\times e\times V}}}}{\frac{h}{\sqrt{2\times m_p\times e\times V}}}=\frac{1}{2\sqrt{2}}$