rydberg equation [ENDORSED]

[KiaraTenorio_14B]

In the lecture notes it is stated that to find the wavelength from moving energy levels on a spectral line that we do En=-(hR/n^2), but then in the hints of the achieve hw it advises us to use the equation 1/lambda=R (1/n^2)-(1/n^2), which equation should we use as each resulted in different steps to achieve the needed answer.

[Jessica Tam 3H]

Hi,
I believe that 1/lambda= (1.0974×10^7)[(1/n^2)-(1/n^2)] was derived from En=-(hR/n^2) because in order to find the difference in energy you subtract the En(final) and En(initial). That would give you the frequency=R [(1/n^2)-(1/n^2)]. The way you get 1/lambda is from c=lambda*frequency. So you would get c/lambda = R [(1/n^2)-(1/n^2)]. Then you would divide the speed of light on both sides leaving you with 1/lambda= (1.0974×10^7)[(1/n^2)-(1/n^2)].

[Miranda Hess 3I]

Yes, the second one is derived from the first one. Based on the lecture, you can use either one but you need to know the first one and what all of the pieces mean in order to fully understand the process of using both equations.

[Benicio Rivera 1F]

1/lambda=R (1/n^2)-(1/n^2) Is derived from the lecture eq. Understand the lecture eq and use 1/lambda=R (1/n^2)-(1/n^2) for the hw.

[Michelle Gong]

Hello!

The second equation is derived from the first equation, meaning that you can use either interchangeably depending on your situation / whatever is easier. 1/lambda comes from c = lambda * frequency --> c/lambda = R [(1/n^2)-(1/n^2)].

[Emily Widjaja 3A]

Hi,

I think the second equation is just derived from the first equation but i personally use the second one more when doing questions.