 # What would be the atomic number of the atom whose 1s electron moves at 0.99c (the speed of light)? analianopolisca 2022-08-07 Answered
What would be the atomic number of the atom (may be hypothetical) whose 1s electron moves at $0.99c$ (the speed of light)?
Quantum mechanics might have an answer, but I do not know the necessary maths to calculate.
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You can get a back of the envelope notion of the energy of a inner-most orbital by just treating the problem as a hydrogen-like atom (not entirely fair and almost certainly a slight over-estimate but at least it is easy). You get
$\begin{array}{r}{E}_{1s}\approx \mathrm{R}\mathrm{y}\ast {Z}^{2}=\left(13.7\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}\right)\ast {Z}^{2}\phantom{\rule{thickmathspace}{0ex}}.\end{array}$
Where $Z$ is the atomic number of the atom in question and $\mathrm{R}\mathrm{y}=13.6\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$ is the Rydberg constant.
If you are asking for a speed of $\beta =0.99$ ($\gamma =7.1$) then you are suggesting an kinetic energy of about $T=\left(\gamma -1\right){m}_{e}{c}^{2}=6.1\left(5.11×{10}^{5}\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}\right)=3.1×{10}^{6}\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$. Which suggests:
$\begin{array}{rl}{Z}^{2}& =\frac{\left(\gamma -1\right){m}_{e}{c}^{2}}{\mathrm{R}\mathrm{y}}\\ & \approx \frac{6.1\left(5.2×{10}^{5}\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}\right)}{13.7\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}}\\ & =2.3×{10}^{5}\\ Z& \approx 480\phantom{\rule{thickmathspace}{0ex}},\end{array}$
give or take a small factor.
Even for $\beta =0.9$ ($\gamma =2.3$) you get $Z\approx 220$.
- For $\beta =0.75$ ($\gamma =1.5$) I find $Z\approx 140$.
- For $\beta =0.65$ ($\gamma =1.3$) I find $Z\approx 110$.
- For $\beta =0.55$ ($\gamma =1.2$) I find $Z\approx 86$.
All of these values are thoroughly relativistic, but as you can see the ultra relativistic regime requires unreasonable heavy nuclei.

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