Answer
$\lambda =0.40 nm$
Work Step by Step
To find the De Broglie Wavelength,
$\lambda = \frac{h}{mv_F}$
Where
$h = 6.626\times 10^{-34} J.s$
$m=9.109\times 10^{-31} kg$
$v_F = 1.82 \times 10^{6} m/s$
So
$\lambda = \frac{6.626\times 10^{-34} J.s}{(9.109\times 10^{-31} kg)(1.82 \times 10^{6} m/s)}$
$\lambda =0.40 nm$
