0
$\begingroup$

In the link, Solving Dirac for one electron atom/ion the energy for the one electron $s_1$ shell is calculated as $$ E=m_{\mu}c^2\left( 1 - \frac{1}{\sqrt{1+\left( \frac{\alpha Z}{n-j-1/2 + \sqrt{(j+1/2)^2-\alpha^2 Z^2}} \right )^2}}\right). $$

I am confused because if $Z=92$, then this is not computable for $j=0$ as $\alpha 92 > 1/2$ and then $ \sqrt{(j+1/2)^2-Z^2\alpha^2}$ is imaginary. Why's that? Is the formula in the link wrong?

Improve this question
$\endgroup$
3
  • $\begingroup$ The Klein Paradox: At Large $Z$ the vacuum is unstable to pair creation in which an electron from the filled Dirac sea can tunnel though the potential barrier and occupy the energy lowest state. $\endgroup$
    – mike stone
    Commented Dec 12, 2021 at 22:34
  • 1
    $\begingroup$ @mikestone Perhaps but there is no such thing as a filled Dirac sea. $\endgroup$
    – my2cts
    Commented Dec 12, 2021 at 23:51
  • $\begingroup$ @ my2cts Why not? It is convenient way to keep track, and as (these days) a condensed matter theorist it is how I have learned to think of things. The argument about infinite electric charge is misleading. In the standard model each generation has $3\times 2/3 $ up quarks, $-3\times 1/3$ down quarks, and $-1$ electon-like particle, Net vacuum charge for each generation is zero. There is nothing that the filled sea gets wrong. $\endgroup$
    – mike stone
    Commented Dec 13, 2021 at 1:36

1 Answer 1

Reset to default
1
$\begingroup$

As far as I understand, $j=0$ is impossible as $l=j\pm\frac{1}{2}$. Look at the definitions of $l$ and $j$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes that makes sense, using this improves the values for other elements as well $\endgroup$
    – Stefan
    Commented Dec 12, 2021 at 23:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.