Answer
$$\lambda = 332.12 pm$$
Work Step by Step
According to De Broglie, wavelength of particle moving at a velocity v is given by
$$\lambda = \frac{h}{mv}$$
where m is the mass of the particle.
Also, $v = \alpha c = \frac{c}{137}$
$$\lambda = \frac{6.626\times 10^{-34}}{9.109\times 10^{-31}
\times \frac{1}{137}\times 3\times 10^{8} } = 332.17\times 10^{-12}m$$
$$\lambda = 332.12 pm$$
