The de-brogile wavelength of an electron in the ground state of hydrogen atom is:

$[K.E.=13.6eV;1eV=1.602\times {10}^{-19}J]$

The correct option is C. $0.3328nm$.

We know that,

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

$\Rightarrow p=\sqrt{2mKE}$

The de-Brogile wavelength is given by,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mKE}}$

where $h=6.626\times {10}^{-34}$ is called Planck's constant.

$\Rightarrow \lambda =\frac{6.626\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 13.6\times 1.6\times {10}^{-19}}}$

$=\frac{6.626\times {10}^{-34}}{19.90\times {10}^{-25}}$

$=3.329\times {10}^{-10}$

$=0.3329nm$.