• # 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

 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].