b
\(\frac{1}{2}m{\upsilon ^2}\, = \,\,eV\,\,\,\,\therefore \,\,{m^2}{\upsilon ^2}\, = \,\,{p^2}\, = \,\,2\ meV\,\,\,\)

\(\therefore \,p\,\, = \,\,\,\frac{h}{{\sqrt {2meV} }}\,\,\,\,\therefore \,\,p\,\, \propto \,\,\frac{1}{{\sqrt {me} }}\,\,\,\,\therefore \,\,\frac{{{\lambda _e}}}{{{\lambda _p}}}\,\, = \,\,\sqrt {\frac{{{m_p}{e_p}}}{{{m_\alpha }{e_\infty }}}} \)

હવે \({{\text{m}}_\alpha }{\text{  =  4mp, e}}\alpha {\text{  =  2}}{{\text{e}}_{\text{p}}}\) હોવાથી 

\(\therefore \,\,\,\frac{{{\lambda _\infty }}}{{{\lambda _p}}}\,\, = \,\,\sqrt {\frac{{{m_p}{e_p}}}{{4{m_p}\,\, \times \,\,2{e_p}}}} \,\,\, = \,\,\,\frac{1}{{\sqrt 8 }}\,\,\,\)

\(\therefore \,\,\,\frac{{{\lambda _\infty }}}{{{\lambda _p}}}\,\, = \,\,\,\frac{1}{{2\sqrt 2 }}\,\,\,\,\therefore \,\,\frac{{{\lambda _\infty }}}{{{\lambda _0}}}\,\, = \,\,\,\frac{1}{{2\sqrt 2 }}\,\,\,\,\,\therefore \,\,\,{\lambda _\infty }\, = \,\,\,\frac{{{\lambda _0}}}{{2\sqrt 2 }}\)