Answer
$1.914\times 10^{-21}$ $kg.m/s$
Work Step by Step
For sodium ion, the relation between momentum $p$ and energy $E$ is given by
$E=\frac{p^2}{2m}$
or, $p=\sqrt {2mE}$ ......................$(1)$
where $m$ is the mass of a sodium ion.
If singly charged sodium ions are accelerated through a potential difference $V$, then $E=eV$, where $e$ is the charge of an electron.
Thus eq. $1$ becomes
$p=\sqrt {2meV}$
$\text{Mass of a sodium ion}(m)=\frac{\text {Molar mass of sodium}}{\text{Avogadro's number}}$
or, $m=\frac{22.9898}{6.022\times 10^{23}}$ $g$
or, $m\approx 3.817\times 10^{-23}$ $g$
or, $m\approx 3.817\times 10^{-26}$ $kg$
Given, $V=300$ $V$
Therefore, the momentum acquired by a sodium ion is given by
$p=\sqrt {2meV}$
or, $p=\sqrt {2\times 3.817\times 10^{-26}\times 1.6\times 10^{-19}\times300}$
or, $p\approx 1.914\times 10^{-21}$ $kg.m/s$
