If the velocity of hydrogen molecule is $5\times {10}^4\mathrm{cm}{\sec }^{-1},$ then its de-Broglie wavelength is

According to de-Broglie

$\begin{array}{l}\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.626\times {10}^{-27}\mathrm{erg}.\sec }{\frac{2}{6.023\times {10}^{23}}\times 5\times {10}^4\mathrm{cm}/\sec }\\ \mathrm{Since} \mathrm{mass} \mathrm{of} \mathrm{a} \mathrm{Hydrogen} \mathrm{molecule} =\frac{2}{6.023\times {10}^{23}}\\ =\frac{6.62\times {10}^{-27}\times 6.023\times {10}^{23}}{2\times 5\times {10}^4}\mathrm{cm}\\ =4\times {10}^{-8}\mathrm{cm}=4{\mathrm{A}}^{\mathrm{o}}.\end{array}$