Question

A proton when accelerated through a potential difference of V volt has a wavelength $\lambda$ associated with it. An $\alpha$ - particle in order to have the same wavelength $\lambda$ must be accelerated through a p.d. of

• A. V/8 volt
• B. V/4 volt
• C. V volt
• D. 2V volt

Answer 1

Answer: (A)

For proton, ${\lambda }_p=\frac{0.286\times {10}^{-10}}{\sqrt{\nu }}m$
where V is p.d
For $\alpha -particle,{\lambda }_{\alpha }=\frac{0.101\times {10}^{-10}}{\sqrt{\nu }}m$

so, keeping same $\alpha$ for proton and $\alpha -partside$

$1=\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{0.286\times \sqrt{V_{\alpha }}}{0.101\times \sqrt{V_P}}$

$0.101\times \sqrt{V_P}=0.286\times \sqrt{V_{\alpha }}$
$(0.101{)}^2\times V_P=(0.286{)}^2{V}_{\alpha }$

$\frac{(0.101{)}^2\nu }{(0.286{)}^2}=V_{\alpha }$

$(\because V_P=V)$

$\Rightarrow V_{\alpha }=\frac{V}{8}volt$

So, the answer is option (A).