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I have read that hydrogen does not produce X-rays as the energy difference between its shells is small.

So if this is true what are the elements that do not produce X-rays and how was Moseley able to plot the frequency of the radiation for such elements?

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  • $\begingroup$ Is the statement in the quote block yours, or is that quoted from something? $\endgroup$ – Bill N Apr 19 at 19:26
  • $\begingroup$ It was an answer to a question $\endgroup$ – Asher2211 Apr 19 at 19:28
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So-called "K-shell" x-rays are produced when electrons fall from outer electron orbitals to the innermost. (The "L-shell" x-rays are produced when a vacancy is filled in the second-innermost shell.)

An electron falling to the hydrogen ground state emits ultraviolet radiation in a spectrum first observed by Lyman. That is, the physics of the hydrogen Lyman series is mostly the same as the physics of K-shell x-rays in heavy elements. Likewise the physics of the Balmer series of visible-light hydrogen transitions is analogous to L-shell x-rays in heavy nuclei.

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  • $\begingroup$ Why do we get frequency of radiation=$0$ for hydrogen when we apply Mosley's formula? $\endgroup$ – Asher2211 Apr 19 at 20:38
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    $\begingroup$ Moseley’s law is an approximation derived for heavy nuclei. It predicts the transition energy is proportional to $(Z-b)^2$ rather than $Z^2$. The constant $b$ reflects “screening” of the nuclear charge by the inner electrons (note from the link that $b_K \approx 1$ and $b_L\approx 8$). But a neutral hydrogen atom doesn’t have any inner electrons to screen the proton’s charge, so the Moseley approximation is bad. $\endgroup$ – rob Apr 20 at 0:13

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