Answer
$3.464\times 10^{-13}$ $m$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$
where $h$ is the planck's constant and $p$ is the momentum of the moving particle.
From the previous section of this problem, we have calculated the momentum of a singly charged sodium ion:
$p=1.914\times 10^{-21}$ $kg.m/s$
Thus, the de Broglie wavelength of the sodium ion is given by
$\lambda=\frac{6.63\times 10^{-34}}{1.914\times 10^{-21}}$ $m$
or, $\lambda\approx 3.464\times 10^{-13}$ $m$
