Question

An $\alpha$ - particle and a proton are accelerated from rest by a potential difference of $200\text{V}$. After this, their de Broglie wavelengths are ${\lambda }_{\alpha }$ and ${\lambda }_p$ respectively. The ratio $\frac{{\lambda }_p}{{\lambda }_{\alpha }}$ is :

• A. $8$
• B. $2.8$
• C. $3.8$
• D. $7.8$

Answer 1

The correct option is B $2.8$

We know that,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mqV}}$

$\therefore \frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}}=\sqrt{\frac{4\times 2}{1\times 1}}=2.8$