question_answer

An \[\alpha \]-particle and a proton are accelerated through the same potential difference. Calculate the ratio of linear momenta acquired by the two.                       

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, \[p=m\upsilon =\sqrt{2mqV}\] \[\therefore \]  \[\frac{{{p}_{\alpha }}}{{{p}_{p}}}=\sqrt{\frac{2{{m}_{\alpha }}{{q}_{\alpha }}V}{2{{m}_{p}}{{q}_{p}}V}}\] \[=\sqrt{\frac{2\times 4{{m}_{p}}\times 2e\times V}{2\times {{m}_{p}}\times e\times V}}=\mathbf{2}\sqrt{\mathbf{2}}\mathbf{:1}\]