The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature ${27}^{\mathrm{o}}\mathrm{C}$ and${127}^{\mathrm{o}}\mathrm{C}$ respectively is $\sqrt{\frac{\mathrm{x}}{\mathrm{y}}}$ . Find  x+y

de-Broglie wavelength $\mathrm{\lambda }=\frac{\mathrm{h}}{{\mathrm{mv}}_{\mathrm{rms}}}$ , rms velocity of a gas particle at the given temperature (T) is given as 

$\frac{1}{2}{\mathrm{mv}}_{\mathrm{rms}}^2=\frac{3}{2}\mathrm{kT}\Rightarrow {\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{kT}}{\mathrm{m}}}\Rightarrow {\mathrm{mv}}_{\mathrm{rms}}=\sqrt{3\mathrm{mk}\mathrm{T}}$

$\therefore \mathrm{\lambda }=\frac{\mathrm{h}}{{\mathrm{mv}}_{\mathrm{rms}}}=\frac{\mathrm{h}}{\sqrt{3\mathrm{mkT}}}$

$\Rightarrow \frac{{\mathrm{\lambda }}_{\mathrm{H}}}{{\mathrm{\lambda }}_{\mathrm{He}}}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{He}}{\mathrm{T}}_{\mathrm{He}}}{{\mathrm{m}}_{\mathrm{H}}{\mathrm{T}}_{\mathrm{H}}}}=\sqrt{\frac{4(273+127)}{2(273+27)}}=\sqrt{\frac{8}{3}}$