ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: PB on August 24, 2013, 01:18:12 pm
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For example, a hydrogen atom had a groundstate + 4 excited states. if the atom electrons were excited to the 4th excited state, how many possible photon energies could the atom emit on its way down?
I did some random, 4+3+2+1=10 , which happens to be correct! I was wondering if this formula is a legit quick way to solve this question?
because I don't really want to have to count every single possibility and risk missing one!
Thanks,
PB
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It's simple, just remember:
light is emitted when an electron jump from a higher state to lower state.
So all you really have to do, is to count how money possible jumps can an electron make.
Finally, all you have to do is just to calculate the energy difference for each jump.
Hope that helps :)
and please correct me if I made any mistake, thanks.
Also, if its ok, I would really appreciate for a "+" vote.
ops, I didn't see your last sentence, sorry.
your calculation is correct, that is in fact the correct way of calculate how many energy states there is.
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PB, your method works, as long as no two energy level differences are the same. e.g. if 4 -> 3 is the same as 2 -> ground.
asterio, you missed two jumps. 3 -> 1 and 4 -> ground.
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Thanks for the imput Asterio, haha sure you can have a +1.
@socialrhubarb. Good point.