Comparing Electronic Transitions in H and Be3+

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In summary, statement a) is false because the electronic transition from ni=5 to nf=2 in H lies in the visible spectrum while the electronic transition from ni=2 to nf=5 in Be3+ lies in the UV region of the spectrum. Statement b) is false because a photon of the same frequency is not involved in both transitions. And statement c) is true. The energy transitions in hydrogen-like atoms increase as nuclear charge increases, resulting in higher energy transitions. The frequency for a given transition in Be3+ is larger than in H, placing it in the UV region of the spectrum.
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fogvajarash
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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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  • #2
How do the energies (and thus also energy differences) in hydrogen like atoms depend on nuclear charge?
 
  • #3
As nuclear charge increases, shouldn't the energy transitions become higher in energy?
 
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  • #5
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!
 

1. What are electronic transitions in atoms?

Electronic transitions refer to the movement of electrons between different energy states in an atom. These transitions occur when an electron absorbs or emits energy, causing it to move from one energy level to another.

2. How do electronic transitions differ in H and Be3+?

The electronic transitions in H and Be3+ differ in terms of the number of energy levels available for electron movement. H has only one electron, so it can only undergo transitions between its two energy levels. Be3+, on the other hand, has three electrons and therefore has more possible energy levels for transitions.

3. Why is Be3+ used for comparison with H in electronic transitions?

Be3+ is often used for comparison with H because they both have similar atomic structures, with one valence electron in the outer shell. This makes it easier to compare their electronic transitions and understand the underlying principles.

4. What factors influence the energy of electronic transitions in atoms?

The energy of electronic transitions in atoms is influenced by several factors, including the number and arrangement of electrons in an atom, the strength of the electric field, and the type of transition (e.g. from ground state to excited state or vice versa).

5. How are electronic transitions in atoms studied and measured?

Electronic transitions in atoms are studied and measured using spectroscopy techniques. These involve shining light of different wavelengths on an atom and observing the absorption or emission of light at specific wavelengths, which can provide information about the energy levels and transitions in the atom.

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