Question

If ${\lambda }_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100\text{V}$, then the de-Broglie wavelength for $\alpha$ particle accelerated through the same potential difference is -

• A. $2\sqrt{2}{\lambda }_0$
• B. $\frac{{\lambda }_0}{2}$
• C. $\frac{{\lambda }_0}{2\sqrt{2}}$
• D. $\frac{{\lambda }_0}{\sqrt{2}}$

Answer 1

The correct option is C $\frac{{\lambda }_0}{2\sqrt{2}}$

We know that,

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

As the particles are accelerated through the same voltage,

$\lambda \propto \frac{1}{\sqrt{mq}}$

So,

$\frac{{\lambda }_{\alpha }}{{\lambda }_p}=\sqrt{\frac{m_p{q}_p}{m_{\alpha }{q}_{\alpha }}}$

$\Rightarrow \frac{{\lambda }_{\alpha }}{{\lambda }_p}=\sqrt{\frac{1\times e}{4\times 2e}}$

$\Rightarrow {\lambda }_{\alpha }=\frac{{\lambda }_p}{2\sqrt{2}}=\frac{{\lambda }_0}{2\sqrt{2}}$

Hence, option $\left(C\right)$ is the correct answer. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->