Atomic Spectra

[Rose_Malki_3G]

For the equation V = -R((1/n^2)-(1/n^2))
How do we know which n value (the final or initial n value) to put first in the equation?

[Vivian Chang 3L]

Hi!

You can manipulate the equation you mentioned to get v(frequency) = R_H (1/n₂^2 - 1/n₁^2), where R_H is Rydberg's constant, 3.29 x 10^15 s^/1.

In this equation, n₁ is always the higher energy level and n₂ is the lower energy level.

[Anh Trinh 1J]

The first n in the equation would be the final energy level. Dr. Lavelle said in today’s lecture to think of it as the change in energy = final energy - initial energy. Usually in a question it would describe the transition that the electron makes. Ex: “an electron makes a transition from the fourth to the second principal quantum level,” the fourth energy level would be the initial and the second energy level would be the final.

[LeanneBagood_2F]

For the equation V = -R((1/n^2)-(1/n^2))
How do we know which n value (the final or initial n value) to put first in the equation?

Some of our classmates already responded with this but the difference of something is always going to be found by calculating final - initial.
Because of this, the first n in the equation should be n-final whereas the second should be n-initial. Here's an image of Rydberg's equation that I got from Google if my explanation does not make sense!

little disclaimer: i got this image from a quick Google search of "rydberg's equation" so it does not exactly align with the equation that was originally being asked about but it gives the gist of it

[AndrewNguyen_2H]

Hi!

You can manipulate the equation you mentioned to get v(frequency) = R_H (1/n₂^2 - 1/n₁^2), where R_H is Rydberg's constant, 3.29 x 10^15 s^/1.

In this equation, n₁ is always the higher energy level and n₂ is the lower energy level.

Isn't n1 the initial??

[Justin Nguyen 3E]

I think n1 is the initial and also the higher energy level because the electrons emit energy when an excited electron (higher energy level) drops back down to its ground state.