Answer:
If a particle is accelerated through a potential difference V, then \[qV=\frac{1}{2}m{{\upsilon }^{2}}=\frac{{{p}^{2}}}{2m}\] or \[p=\sqrt{2mqV}\] \[\therefore \] \[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mqV}}\] As \[{{\lambda }_{p}}={{\lambda }_{\alpha }}\] or \[\frac{h}{\sqrt{2{{m}_{p}}{{q}_{p}}{{V}_{p}}}}=\frac{h}{\sqrt{2{{m}_{\alpha }}{{q}_{\alpha }}{{V}_{\alpha }}}}\] or \[\frac{{{V}_{p}}}{{{V}_{\alpha }}}=\frac{{{m}_{\alpha }}{{q}_{\alpha }}}{{{m}_{p}}{{q}_{p}}}=\frac{4{{m}_{p}}.2e}{{{m}_{p}}.e}=\mathbf{8}\]
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