When neutral $\ce{Cl}$ atoms are bombarded by high energy photons, causing ejection of electrons, which subshells' ejected electrons will have highest velocity?

I chose $1s$ because I thought more energy is needed to ejected those electrons so they might have higher velocity. My second choice $3s$ is right, but I can't explain why.


1 Answer 1


It would be the $3s$ shell because less energy is required to bump it off course. Due to electron repulsion and the shielding effect, the $e-$ in the $3s$ is more prone to get dislodged. And of course because of this, it will have a higher velocity when it gets bumped off. Another concept to note: $e-$ will have more energy at higher energy levels.

A reason why it wouldn't be the $1s$ shell is because due to the strong pull on the $1s$ electron. Since it is on the first energy level, it is strongly held by the nucleus. Even it were to get dislodged it would move very slow because of the strong pull by the nucleus.

  • $\begingroup$ so it's because 3s electrons after bumped off will have more kinetic energy leftover right? $\endgroup$ May 3, 2015 at 22:47
  • $\begingroup$ Exactly. You figured it out. $\endgroup$
    – Asker123
    May 3, 2015 at 22:48
  • 2
    $\begingroup$ Why not $3p$? Aren't they further out and higher energy than $3s$ electrons? $\endgroup$
    – user213305
    Oct 12, 2017 at 22:01

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