Question

A proton and an $\alpha$-particle are accelerated through the same potential difference. The ratio of their De Broglie wavelength is:

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $2\sqrt{2}$
• D. $2$

Answer 1

Answer: (C)

$2\sqrt{2}$

According to De-broglie,

$\lambda =h/mv$

$\lambda =\frac{h}{\sqrt{2m\times K.E}}$

${\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }\times 2e\times v}}$ (charge on $\alpha =2e$)
${\lambda }_p=\frac{h}{\sqrt{2m_p\times e\times v}}$ (charge on $p=e$)

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

as the mass of an alpha particle is 4 times of proton.

Hence, the correct option is $C$