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 ?
Answer & explanation
Correct answer: option 2
$\text{Wavelength is given by } \lambda = \frac{h}{p} = \frac{h}{\sqrt 2mE} = \frac{h}{\sqrt 2mqV}$
$\frac{\lambda_p}{\lambda_{\alpha}}= \sqrt{ \frac{ m_{\alpha} q_{\alpha}}{ m_{\alpha} q_{\alpha}}} = \sqrt 8$
$ \lambda_p = \sqrt 8 \lambda_{alpha}$