Atomic Spectrum

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Taravat_Lakzian_1C
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Joined: Wed Sep 21, 2016 2:56 pm

Atomic Spectrum

Postby Taravat_Lakzian_1C » Fri Sep 30, 2016 2:30 pm

For question 1.13, it says to use the Rydberg formula for atomic hydrogen to calculate the wavelength of radiation generated by the transition from n=4 to n=2. What is the Rydberg formula?

Daniel_Callos_4I
Posts: 16
Joined: Wed Sep 21, 2016 2:57 pm

Re: Atomic Spectrum

Postby Daniel_Callos_4I » Fri Sep 30, 2016 2:39 pm

The Rydberg equation is a mathematical extension of the Bohr Frequency Condition which allows us to calculate the frequency of photons emitted as photoexcited electrons fall back to their original energy level. In a former post, I gave an informal proof for the Rydberg equation based on the basic statement of the Bohr frequency condition. I have included this information word-for-word from my former post below:

Ep = E2 - E1

where Ep is the energy of the photon, E2 is the energy of the excited electron, and E1 is the energy of the electron before excitation. In concert with our knowledge of E=hv, we can mathematically iterate this phenomenon known as Bohr's Frequency Condition that solves for the frequency of incident light:

hv = (E1 - E2)
v = (E1 - E2)/h

Note that it was experimentally determined that the energy En at a given energy level is equal to -hR/n^2, where h is Planck's constant and R is the Rydberg constant (3.29*10^15 s^-1). The convention is that an electron with no relation to the atom would be at an infinitely high energy level n and thus its energy would approximate 0. Therefore, electrons in orbit are said to have negative energy since their energy is lower than that of a freed electron. Thus we can extend upon Bohr's Frequency condition as follows:

v= (-hR/(n1)^2 - -hR/(n2)^2)/h [replace E with -hR/n^2]
v= -hR(1/(n1)^2 - 1/(n2)^2)/h [factor out -hR]
v=-R(1/(n1)^2 - 1/(n2)^2) [h cancels, we are left with the Rydberg eqn]

Hope this information helps and have a great day!

-Dan Callos


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