Question 1.15

[CalebBurns3L]

Okay so I understand how to get that the two n's are n = 1 and n=3, but how are you supposed to know that it goes from n=1 to n=3 and not the other way around?

[Clement Ng]

This is because in the formula v= R(1/n^2 - 1/N^2), if n was 3 and N was 1, the frequency would be a negative number.

1/9 - 1/1 = negative number.

[Julian Krzysiak 2K]

is just another way of writing

When we convert the equation, the corresponds to , and corresponds to

We know that , the final energy level, and , the initial energy level, because "the electron during the emission of energy"- for an electron to emit energy, it has to move down energy levels. Because the electron is moving down, it must start from n=3, therefore the initial, and must end at the lower energy level, n=1

[CalebBurns3L]

@Julian Krzysiak 3L

No but so that's what's confusing me. I thought it should go exactly as you described, but in the solutions Manual it says that n=1 is initial and n= 3 is the final, which is the opposite of what you said :( and also opposite to what would make sense to me if the electrons are "emitting" energy.

Can anyone explain what's going on here?

Thank you!

[Ethan-Van To Dis2L]

Okay based on my discussion section my TA said that n1 is considered the intial and n2 is the final. You would make the lowest n value (in this case 1) your initial that way the frequency always remains positive, which may explain why your initial n is 1 and your final n is 3. This might be confusing since it would be technically n=3 as your inital and n=1 would be final since it's emitting light. Hoped this helped!

[Cam Bear 2F]

So for this equation, we just order the energy levels to insure it's positive rather than doing final - initial energy level?

[Jessica Jones 2B]

In most problems, it clearly states the order of this change, but if it does not you would use the equation: E= -hR/n^2
Stay away from V= R[1/n1^2 - 1/n2^2] because it will not clearly demonstrate what is truly happening in the problem.