## n1 and n2

$\lambda=\frac{h}{p}$

705340227
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### n1 and n2

In the equation v=R[1/n1^2-1/n2^2], is n1 or n2 the shell that the electron originally starts at before it moves down the shells?

Sachi Sengupta 2C
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### Re: n1 and n2

N1 is basically the final energy level and N2 is the initial energy level. During a UA session, they even suggested writing nf and ni instead so that its less confusing. Hope this helped!

Justin Zhang_1A
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### Re: n1 and n2

For that equation, n2 should be the shell that the electron originally starts at. It's easier to think of it as n(final) - n(initial).

Sonel Raj 3I
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### Re: n1 and n2

n1 is nFinal and n2 is nInitial, which is an easier way of looking at it. Also, for this equation, you have to always make sure the smaller n-value (bigger number overall) is in front, or else it does not work. I have no idea why, but you can always default to the other equation Lavelle gave us if this one doesn't work, lol

Marc Lubman 3B
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### Re: n1 and n2

N2 would be the shell that the electron starts from. It might be more helpful to think of the equation as $\nu =R\left ( \frac{1}{n_{final}^{2}}-\frac{1}{n_{initial}^{2}} \right )$ than just as n1 and n2.

Dylan_Nguyen_2C
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### Re: n1 and n2

For this equation, I like to think of it as v=R(1/nfinal^2 - 1/ninitial^2), meaning the final n-value(the smaller one) comes first in the equation and conversely the initial larger n-value comes second in the equation. Keep in mind that the resulting fraction inside the parenthesis(the number multiplied by R) should be positive. If the fraction is negative then the order of the n-values in the equation is wrong and will lead to an incorrect answer. Hope this helps

Keon Amirazodi 3H
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### Re: n1 and n2

n1 is usually the final energy level and n2 is the initial energy level. But rather than thinking of it as 1 minus 2, also remember that it is final energy level minus initial energy level.

IshanModiDis2L
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### Re: n1 and n2

I would prefer you think about it as initial energy level and final energy level rather than just mathematically as light operates at discrete energy levels. It is basically n(final)- n(initial) and n2 should be the initial shell where it begins and n1 where it ends up.

Savana Maxfield 3F
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### Re: n1 and n2

Not sure if this has already been covered, but is the Rydberg equation only used when an electron moves down energy levels? Or when an electron moves to higher energy levels as well?

Kyle Dizon 3A
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### Re: n1 and n2

It will be easier to remind yourself that it should be final minus the initial.

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### Re: n1 and n2

N2 ia the final while N1 is the initial

Simran Bains 2C
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### Re: n1 and n2

The n1 refers to the final energy level and n2 is the initial energy level.

Jack Kettering 3D
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### Re: n1 and n2

N1 is the final and N2 is the initial energy level of the electron. I find this very confusing so I would suggest memorizing the equation using nf and ni to denote final and initial positions of the electron :)

Andrew Yoon 3L
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Joined: Wed Sep 30, 2020 9:36 pm

### Re: n1 and n2

In this case, n=1 would be the final energy and n=2 would be the initial energy level. In order for it to be less confusing, you can replace n1 and n2 with nfinal and ninitial

205323697
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### Re: n1 and n2

How can we split this equation, if that is even possible?

rhettfarmer-3H
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### Re: n1 and n2

Yeah so n1 is final and n2 is initial. A better way to remember it for me is the overall energy is derived from the equation change in E=E(final)-E(intial). Therefore the n1 comes from the final and by the way dont use the ryberg equation makes it harder on yourself.