A Li^2+ ion of mass 7.0 u, initially at rest with its electron in the n=3 level, makes a transition directly to the n=1 level by emitting a single…

For $\mathrm{Li^{2+}}$, $Z=3$, so $E_n=-13.6\times9/n^2\ \mathrm{eV}$: $E_3=-13.6\ \mathrm{eV}$ and $E_1=-122.4\ \mathrm{eV}$. The emitted photon has energy $E=E_3-E_1=108.8\ \mathrm{eV}=1.741\times10^{-17}\ \mathrm{J}$. A photon of energy $E$ carries momentum $p=E/c$, and since the ion was at rest, momentum conservation gives it an equal and opposite momentum, so $Mv=E/c$. With $M=7.0\times1.66\times10^{-27}=1.162\times10^{-26}\ \mathrm{kg}$, $v=\dfrac{E}{Mc}=\dfrac{1.741\times10^{-17}}{1.162\times10^{-26}\times3.0\times10^{8}}=4.99\ \mathrm{m\,s^{-1}}$. The recoil kinetic energy is of order $10^{-25}\ \mathrm{J}$, utterly negligible compared with the photon energy, so ignoring it in the photon energy is justified. Equivalently the photon momentum is $E/c=5.8\times10^{-26}\ \mathrm{kg\,m\,s^{-1}}$, which the ion of mass $1.16\times10^{-26}\ \mathrm{kg}$ takes up in the opposite direction, giving the same speed.