A proton and an alpha-particle are accelerated using the same potential difference. How are the de-broglie wavelengths \(\lambda\)_p and \(\lambda\)_a related to each other ?

\(\lambda\)_p = \(\sqrt {8}\) \(\lambda\)_a

$\text{Wavelength is given by } \lambda = \frac{h}{p} = \frac{h}{\sqrt 2mE} = \frac{h}{\sqrt 2mqV}$

$\frac{\lambda_p}{\lambda_{\alpha}}= \sqrt{ \frac{ m_{\alpha} q_{\alpha}}{ m_{\alpha} q_{\alpha}}} = \sqrt 8$

$ \lambda_p = \sqrt 8 \lambda_{alpha}$