Question

A proton, when accelerated through a potential difference of $V\text{volts}$ has a wavelength $\lambda$ associated with it. If an $\alpha -$ particle is to have the same wavelength $\lambda ,$ it must be accelerated through a potential difference of $V/n\text{volts}.$ The value of $n$ is

Answer 1

Given, ${\lambda }_p={\lambda }_{\alpha }=\lambda$

The de-Broglie wavelength associated with a particle of charge $\left(q\right)$ and mass $\left(m\right)$ accelerated through a potential difference $V$ is given by,

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

$V=\frac{h^2}{2mq{\lambda }^2}[\because \lambda =\text{same for both}]$

$\Rightarrow V\propto \frac{1}{mq}$

$(\because m_{\alpha }=4m_p;q_{\alpha }=2q_p$)

$\therefore \frac{V_{\alpha }}{V_p}=\frac{m_p}{m_{\alpha }}\times \frac{q_p}{q_{\alpha }}=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$

$\therefore V_{\alpha }=\frac{V}{8}$

Hence, $n=8$