If the velocity of hydrogen molecule is 5 x 10⁴ cm sec^-1 then its de-Broglie wavelength is

According to de-Broglie
$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.62\times {10}^{-27}\text{ erg.sec }}{\frac{2}{6.023\times {10}^{23}}\times 5\times {10}^4\mathrm{cm}/\sec }$
$=\frac{6.62\times {10}^{-27}\times 6.023\times {10}^{23}}{2\times 5\times {10}^4}\mathrm{cm}=4\times {10}^{-8}\mathrm{cm}=4\text{Å}$