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I think it must have to do with the fact that an incident photon with a sufficiently high energy will ionize the hydrogen atom (i.e. the electron will be ejected). Therefore it won't show in the emission spectrum. Is this correct?

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    $\begingroup$ Kind of, but backwards. The well is only so deep, so an electron dropping from right at the continuum down to the 1s state can only release 13.6eV. $\endgroup$ – Jon Custer Aug 27 '16 at 20:09
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The closest an e- can get to nucleus is the 1st, it can't get closer and so can't get to lower E. If an e- is beyond the influence of the nucleus and falls to 1s it can't fall any further (drop in E any more ). This represents the largest E emission. Once an e- is beyond the influence of the nucleus it doesn't matter how much further it goes it is still beyond its influence and hence the energy of the transition doesn't increase

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    $\begingroup$ Can you please at least spell out stuff fully? $\endgroup$ – orthocresol Aug 28 '16 at 11:49
  • $\begingroup$ The max E emission is the energy an e- releases upon falling to the 1s from a distance so far away from the nucleus that nucleus has no hold. If an e- is further away makes no difference. Can't make it simpler than that. If you don't understand please be more specific about what you don't get $\endgroup$ – user33810 Aug 28 '16 at 11:56

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