According to de-Broglie's equation,

$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}$

where $\mathrm{\lambda }=$ wavelength of particle

$\mathrm{h}=$ Planck's constant

$\mathrm{p}=\mathrm{mv}=$ momentum of particle

Therefore, $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$

$=\frac{6.63\times {10}^{-34}}{11.043\times {10}^{-26}\times 6.0\times {10}^7} \mathrm{m}$

$=1.0006\times {10}^{-16} \mathrm{m}$