Answer:
The kinetic energy gained by a particle when accelerated through potential difference \[V\] is \[\frac{1}{2}m{{\upsilon }^{2}}=qV\] or \[{{m}^{2}}{{\upsilon }^{2}}=2mqV\] \[\therefore \] Momentum, \[p=m\upsilon =\sqrt{2mqV}\] \[\therefore \] \[\frac{{{p}_{\alpha }}}{{{p}_{p}}}=\sqrt{\frac{2{{m}_{\alpha }}{{q}_{\alpha }}V}{2{{m}_{p}}{{q}_{p}}V}}\] \[=\sqrt{\frac{2\times 4{{m}_{p}}\times 2e\times V}{2\times {{m}_{p}}\times e\times V}}=\mathbf{2}\sqrt{\mathbf{2}}\mathbf{:1}\]
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