# What would be the atomic number of the atom whose 1s electron moves at 0.99c (the speed of light)?

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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Isaias Archer
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.