Rydberg Formula

$c=\lambda v$

Michelle Pham 1G
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Joined: Fri Sep 25, 2015 3:00 am

Rydberg Formula

For Hw 1.13 (6th edition)/Hw 1.7 (5th edition), it the Rydberg formula is to be used for the transition from n= 4 to n =2. $v (frequency)= R(\frac{1}{(n1)^2)}-\frac{1}{(n2)^2})$

I used 4 as n1 and 2 as n2 because it was listed in that order. I checked the Solutions Manual and 2 was used for n1 and 4 for n2. Could someone please explain how you can figure out which number to use for n1 and which number to use for n2 if it is not explicitly stated?

Chem_Mod
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Re: Rydberg Formula

There are a few ways to think about why n2 should be 4 and n1 should be 2. Mathematically, if you plugged in the values in the opposite order (i.e. n1 = 4, n2 = 2) you would arrive at a negative value for frequency, which can not be true as you can not have negative cycles or negative seconds (frequency being cycles/second). Another way to keep n1 and n2 straight is to refer to equation 2 on page 7 of the text book where n2 is defined to be n1 + 1, n1 + 2, ... so n2 must be the larger of the two values. Hopefully one of the these two methods helps clarify your question.

Chem_Mod
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Re: Rydberg Formula

For the Rydberg Equation, the condition is that n2> n1
therefore, since 4 is greater than 2, 4 would be plugged in for n2 and 2 would be plugged in for n1
Also, to gain an understanding of why this condition is so, we can think about frequency. You will want to end with a positive value for your frequency. Therefore, the number plugged in for n2 must be greater than n1.

Daniella Ching 4C
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Re: Rydberg Formula

When do you need to use the Rydberg Formula?

Rex Lee 1C
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Re: Rydberg Formula

The Rydberg Formula arranged for frequency, as stated in the question, can be used whenever the problem asks for the energy/wavelength/frequency of the photon released when an electron falls/is excited from one energy level to another. After plugging in the values for the energy levels, you get the frequency. Energy can subsequently be found by multiplying frequency by the Planck constant, or you can solve for wavelength by dividing "c" by the frequency.

However, you can also use the method that involves calculating the energy for each electron level separately (E=-hR/n^2) then simply calculate Efinal-Einitial for the energy of the resulting photon

Heather Lindsay 1H
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Re: Rydberg Formula

Where does the value for the Rydberg constant come from? In other words, why is it 3.29x10^15 Hz?

Kimberly Wagas 1H
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Re: Rydberg Formula

Can the Rydberg formula only be used for transitions between energy levels of a single hydrogen atom with only one electron? If so, are we expected to know how to calculate the wavelength of light for atoms with multiple electrons?

Erin Min
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Joined: Fri Sep 25, 2015 3:00 am

Re: Rydberg Formula

@HeatherLindsay1H The Rydberg constant is the lowest energy photon, or longest wavelength emitted from the hydrogen atom.