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Author Topic: Atomic transition calculation  (Read 715 times)

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owlpower

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Atomic transition calculation
« on: November 23, 2018, 02:41:01 AM »

I have 3 main concerns regarding the question below.

1) I don't understand the equation that was introduced in the question below.

I can't seem to derive it(out of curiosity), with all the equations I know. E=hf, 1/λ = R(1/n1^2 - 1/n2^2) & λ = c/f

How did the Z get introduced inside? Furthermore, why is (1/n1^2 - 1/n2^2) at the denominator? I thought that it should be in the numerator instead.

2) What do they mean by "...for Hydrogen like atoms..."? I thought they were talking about O7+ there.

3) I don't really understand what the question is getting at and my attempt at the question lead me to deduce that the first 2 transitions are in the visible light spectrum while the last transition(1->4) is in UV spectrum.
I have a feeling I'm getting wrong somewhere though. I took Z to be 1 for all cases (because the first transition number is 1) and I placed (1/n1^2 - 1/n2^2) in the numerator.

Any tips is much appreciated, thank you very much!
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Borek

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Re: Atomic transition calculation
« Reply #1 on: November 23, 2018, 03:20:19 AM »

(1/n12 - 1/n22) is in the numerator, you are misreading a=b/c*d ([itex]a = \frac b c \times d[/itex]) for a=b/(c*d) ([itex]a = \frac b {c \times d}[/itex]).

Quote
What do they mean by "...for Hydrogen like atoms..."? I thought they were talking about O7+ there.

Google "hydrogen like atom" (O7+ counts as one). It will also tell you why you are wrong here:

Quote
I took Z to be 1 for all cases
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Enthalpy

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Re: Atomic transition calculation
« Reply #2 on: November 23, 2018, 11:20:59 AM »

Z is the number of protons, 8 for oxygen.

If there is one single electron, we can still compute the orbital energies. At identical orbital size, the energy would be Z times bigger, but they also shrink by Z in al directions, so the energy is Z2 times bigger.
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owlpower

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Re: Atomic transition calculation
« Reply #3 on: November 23, 2018, 10:14:52 PM »

Thank you for the tips everyone, I managed to get my answers to be in the power of 10^-9 m for the wavelength and thus concluded that the transitions occur in the ultraviolet range, which makes sense.
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