Question

A proton accelerated through a potential V has de-Broglie wavelength $\lambda$. Then the de-Broglie wavelength of an $\alpha$-particle, when accelerated through the same potential V is :

• A. $\frac{\lambda }{2}$
• B. $\frac{\lambda }{\sqrt{2}}$
• C. $\frac{\lambda }{2\sqrt{2}}$
• D. $\frac{\lambda }{8}$

Answer 1

The correct option is C $\frac{\lambda }{2\sqrt{2}}$

As de-Broglie wavelength is given by $\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2m(K.E)}}$

$\lambda =\frac{h}{\sqrt{2mqV}}$ where $v=$ accelerating potential

$q=$charge

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

$\alpha ={}^{4}H{e}^{++}$

$\Rightarrow r{q}_{\alpha }=2q_p$

$m_{\alpha }=4m_p$

${\lambda }_{\alpha }$ at same accelerating potential

${\lambda }_{\alpha }=\frac{h}{\sqrt{2\times 4m_p\times 2q_p\times v}}$

${\lambda }_{\alpha }=\frac{h}{\sqrt{2m_p{q}_{p}v}\times \sqrt{8}}=\frac{\lambda p}{2\sqrt{2}}$

Given ${\lambda }_p=\lambda$

$\Rightarrow {\lambda }_{\alpha }=\frac{\lambda }{2\sqrt{2}}$.