An α– particle and a deuteron $\left({\mathrm{H}}_1^2\right)$ are accelerated from rest by a potential difference of 100V. If λ₁ and λ₂ are their respective de – Broglie wavelengths then $\frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2}$ is

If a charged particle is accelerated by a potential difference of V it's kinetic energy is equal to qV

$\mathrm{K}.\mathrm{E}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}=\mathrm{qV}$

$\mathrm{p} \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} \mathrm{p}=\sqrt{2\mathrm{mqV}}$

$\mathrm{de} \mathrm{Broglie} \mathrm{wavelength} \mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mqV}}}$

$\mathrm{so} \mathrm{\lambda } \mathrm{is} \mathrm{inversely} \mathrm{proportional} \mathrm{to} \sqrt{\mathrm{mq}}$

$\frac{{\mathrm{\lambda }}_{\mathrm{\alpha }}}{{\mathrm{\lambda }}_{\mathrm{D}}}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{D}}{\mathrm{q}}_{\mathrm{D}}}{{\mathrm{m}}_{\mathrm{\alpha }}{\mathrm{q}}_{\mathrm{\alpha }}}}=\sqrt{\frac{2}{4}\frac{1}{2}}=1:2$