• # question_answer A proton and a-particle are accelerated through the same potential difference. The ratio of wavelength of the proton to that of an a-particle is: A)  $2\sqrt{2}:1$           B)  $1:2\sqrt{2}$C)  $2:1$                         D)  none of these

The relation for energy is given by $E=\frac{1}{2}m\,{{V}^{2}}$             or         $\sqrt{2mE}=m\,V$ and relation for wavelength is given by             $\lambda =\frac{h}{m\upsilon }=\frac{h}{\sqrt{2m\,E}}$ Now for proton    ${{\lambda }_{p}}=\frac{h}{\sqrt{2\,P\,E}}$        ?..(i) and for a particle, ${{\lambda }_{\alpha }}=\frac{h}{\sqrt{2\times 4m\times 2E}}$    ?.(ii) (mass of $\alpha$ - particle is 4 times to that of proton) From equation (i) and equation (ii), we get             $\frac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\frac{h}{\sqrt{2m\,E}}\times \frac{\sqrt{16m\,E}}{h}$ $\Rightarrow$               $\frac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\frac{2\sqrt{2}}{1}$ Hence        ${{\lambda }_{p}}:{{\lambda }_{\alpha }}=2\sqrt{2}:1$