Chemistry - Atomic Structure Question with Solution | TestHub

Using de Broglie equation$\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{q}\mathrm{m}\mathrm{V}}}$

$\mathrm{\lambda }=$wavelength, q = charge, m = mass, V = Potential and h = Plank constant

${\mathrm{\lambda }}_{\mathrm{a}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{q}}_{\mathrm{a}}{\mathrm{m}}_{\mathrm{a}}\mathrm{V}}}$${\mathrm{\lambda }}_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{q}}_{\mathrm{p}}{\mathrm{m}}_{\mathrm{p}}\mathrm{V}}}$

${\mathrm{m}}_{\mathrm{a}}=4{\mathrm{m}}_{\mathrm{p}},{\mathrm{q}}_{\mathrm{a}}=2{\mathrm{q}}_{\mathrm{p}}$

$\frac{{\mathrm{\lambda }}_{\mathrm{a}}}{{\mathrm{\lambda }}_{\mathrm{p}}}=\sqrt{\frac{{\mathrm{q}}_{\mathrm{p}}{\mathrm{m}}_{\mathrm{p}}}{{\mathrm{q}}_{\mathrm{a}}{\mathrm{m}}_{\mathrm{a}}}}=\sqrt{\frac{1}{2}.\frac{1}{4}}=\sqrt{\frac{1}{8}}$

$\frac{{\mathrm{\lambda }}_{\mathrm{a}}}{{\mathrm{\lambda }}_{\mathrm{p}}}=\frac{1}{2\sqrt{2}}$