question_answer

A proton, when accelerated through a potential difference of V 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

A) V/8 volts

B) V/4 volts          

C) 4V volts

D) 8V volts

Correct Answer: A

Solution :

The wavelength associated with a particle of charge q, mass m and accelerated through a potential difference V is given by

\[\lambda =\frac{h}{\sqrt{2mqV}}\,\,or\,\,V=\frac{{{h}^{2}}}{2mq{{\lambda }^{2}}}\] for proton : \[V=\frac{{{h}^{2}}}{2{{m}_{p}}}{{q}_{p}}{{\lambda }^{2}}\] for \[\alpha \]- particle \[V'=\frac{{{h}^{2}}}{2{{m}_{\alpha }}{{q}_{\alpha }}{{\lambda }^{2}}}\]

\[(\because {{m}_{\alpha }}=4{{m}_{p}}\,\,and\,\,{{q}_{\alpha }}=2{{q}_{p}})\]

Thus\[V'=V/8\]. Hence the correct choice is [A].