Question

A proton and an alpha particle are accelerated through the same potential difference. The ratio of wavelengths associated with proton and alpha particle respectively is :

• A. $1:2\sqrt{2}$
• B. $2:1$
• C. $2\sqrt{2}:1$
• D. $4:1$

Answer 1

The correct option is C $2\sqrt{2}:1$

We know, $\frac{1}{2}m{v}^2=eV$

$mv=\sqrt{2meV}$

$wavelength,\lambda =\frac{h}{mv}$

${\lambda }_p=\frac{h}{\sqrt{2m_p{q}_{p}V}}$

${\lambda }_p=\frac{h}{\sqrt{2meV}}$

${\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }V}}$

$=\frac{h}{\sqrt{2\times 4m\times 2e\times V}}$

$\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{2\sqrt{2}}{1}$

So, the answer is option (C).