question_answer

Calculate the ratio of the accelerating potential required to accelerate (i) a proton and (ii) an \[\alpha \] -particle to have the same de-Broglie wavelength associated with them. [Given : Mass of proton \[=1.6\times {{10}^{-27}}\] kg ; Mass of \[\alpha \] -particle\[=6.4\times {{10}^{-27}}\] kg]             

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            \[\frac{{{V}_{p}}}{{{V}_{\alpha }}}=\frac{{{m}_{\alpha }}{{q}_{\alpha }}}{{{m}_{p}}{{q}_{p}}}=\frac{4{{m}_{p}}.2e}{{{m}_{p}}.e}=\mathbf{8}\]