Question

A proton is accelerated from rest through a potential difference of V volts and has a wavelength $\lambda$ associated with it. An alpha particle in order to have the same wavelength must be accelerated from rest through a potential difference of:

• A. $\frac{V}{4}volt$
• B. $\frac{V}{2}volt$
• C. $\frac{V}{16}volt$
• D. $\frac{V}{8}volt$

Answer 1

The correct option is D $\frac{V}{8}volt$

$\lambda =\frac{h}{\sqrt{2mqV}}\lambda \alpha \frac{1}{\sqrt{mqV}}\frac{{\lambda }_p}{{\lambda }_a}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }{V}_{\alpha }}{m_{\alpha }{q}_{\alpha }{V}_{\alpha }}}=1\frac{4\times 2\times V_{\alpha }}{1\times 1\times V}=1V_{\alpha }=\frac{V}{8}$