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Topic: calculate the wavelengths for the three  (Read 4634 times)

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Offline jokerboy111

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calculate the wavelengths for the three
« on: October 06, 2006, 05:53:34 PM »
I'm having a bit of trouble with this question. Could someone also explain what the Reidtberg constant is with an infinity sign as a subscript?

A one-electron atom (Be) has the following energy levels:

E1 = -3.49 x 10-17 J
E2 = -8.72 x 10-18 J
E3 = -3.88 x 10-18 J
E4 = -2.18 x 10-18 J
E5 = -1.39 x 10-18 J
E6 = -9.69 x 10-19 J



The spectroscopic series observed for the hydrogen atom are named for
the scientists who discovered them: Balmer, Paschen, Lyman, etc. Within
each series, the initial state of the atom is constant. For the “hydrogen-like”
atom that is of interest here, please calculate the wavelengths for the three
longest wavelength transitions in its Balmer series.


Thanks everyone!!
« Last Edit: October 08, 2006, 01:14:37 PM by Mitch »

Offline jokerboy111

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Re: Alright kids let's give this a try...
« Reply #1 on: October 06, 2006, 05:58:57 PM »
am i gonna violate some rule or something??

Online Borek

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Re: Alright kids let's give this a try...
« Reply #2 on: October 06, 2006, 06:24:35 PM »
am i gonna violate some rule or something??

You better read them (see top of the page).
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Offline Yggdrasil

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Re: Alright kids let's give this a try...
« Reply #3 on: October 06, 2006, 06:54:30 PM »
Quote from: The Rules
4. Please show that you've at least attempted the problem.  We don't mind helping you solve problems but we are ethically opposed to doing homework for you. Violators will have their topic deleted or locked, and subject to banning.

Anyway, I'lll give you some guiding questions to help you with answering the question:

1)  The question asks for the three longest wavelenth transitions.  How do these transitions rank in energy?  Are they the three lowest energy transitions or three highest energy transitions?

2)  The Balmer series consists of transitions from one shell to which specific shell of the hydrogen atom (i.e. what principle quantum number)?

Offline jokerboy111

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Re: Alright kids let's give this a try...
« Reply #4 on: October 06, 2006, 07:53:42 PM »
reply to above

2. is it "2"?

Offline Yggdrasil

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Re: Alright kids let's give this a try...
« Reply #5 on: October 07, 2006, 12:22:11 PM »
Yup.  So, how would you find the energies of the Balmer serries transitions in your one-electron beryllium?

Offline jokerboy111

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Re: Alright kids let's give this a try...
« Reply #6 on: October 07, 2006, 03:45:57 PM »
Im honestly gonna guess and say to use the E=(Z^2)(h)(R)/n^2  formula. Im so lost on this topic

Offline Yggdrasil

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Re: Alright kids let's give this a try...
« Reply #7 on: October 08, 2006, 12:22:33 AM »
I think you are overthinking the problem.  The formula you wrote allows you to calculate the energy of an electron in a given orbital.  However, they give you these energies in the problem, so there is no need to calculate them.  So, when an electron transitions from a higher energy orbital to a lower energy orbital, the energy of the photon released by the transition is equal to the difference in energies of the two orbitals.  For example, if an electron falls from an n = 6 orbital to the n=1 orbital, the photon released has an energy of E = E6 - E1.

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