Comparing Electronic Transitions in H and Be3+

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SUMMARY

The discussion focuses on the electronic transitions in hydrogen (H) and beryllium ion (Be3+), specifically comparing the transition from ni=5 to nf=2 in H with the transition from ni=2 to nf=5 in Be3+. It is established that statement a) is true, as the photon involved in the H transition is in the visible spectrum, while the Be3+ transition photon lies in the UV region. Statement b) is false because a photon of the same frequency is not involved in both transitions; it is emitted in H and absorbed in Be3+. The conclusion is that the correct answer is c), as all statements are not true.

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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?
 
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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!
 

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