Comparing Electronic Transitions in H and Be3+

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Homework Statement


Which statement(s) is (are) FALSE in comparing the electronic transition from ni=5 to nf=2 in H with that of electronic transition from ni = 2 to nf = 5 in Be3+:

a) The photon involved in the H transition lies in the visible spectrum while that involved in the Be3+ transition lies in the UV region of the spectrum.
b) A photon of the same frequency is involved in both transitions, but it is emitted the H electron transition and absorbed in the Be3+ transition.
c) All of the above.

Homework Equations





The Attempt at a Solution


I've discarded b) as it is obviously false. However, i think the final answer it's c), but apparently statement a) is true. I know that the H transition corresponds to the visible light region, but why does the Be3+ one lie in the UV region? (if it is from n = 2?) Or how do transitions in absorption of photons work?
 
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How do the energies (and thus also energy differences) in hydrogen like atoms depend on nuclear charge?
 
As nuclear charge increases, shouldn't the energy transitions become higher in energy?
 
DrDu said:
Yes, exactly. Confer the Rydberg formula for hydrogen like ions:
http://en.wikipedia.org/wiki/Rydberg_formula
So how much larger is the frequency for a given transition in Be3+ than in H?
I'm extremely sorry for not replying! I had my exam the last day and i finally realized what was wrong. I did some calculations with the formula, plugging in the results and i got a result of 42nm (i really don't remember). As visible light is from 400nm-800nm, then it would be in the UV region of the spectrum!