## Use of Rydberg formula [ENDORSED]

$E_{n}=\frac{h^{2}n^{2}}{8mL^{2}}$

Rob_Carter_1L
Posts: 10
Joined: Wed Sep 21, 2016 2:59 pm

### Use of Rydberg formula

During lecture, Professor Lavelle mentioned that he prefers not to use the Rydberg formula, rather E= hR/(n1)^2 - hR/(n2)^2. Is this what we should use on the test?

paulapedrani
Posts: 27
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Use of Rydberg formula  [ENDORSED]

My TA said we could use either one, but since you can get partial credit in the quizzes, he recommended we use the one Lavelle was talking about so that there are less chances of making a mistake and more chances of getting partial credit if we do!

An Dang 3F
Posts: 20
Joined: Fri Sep 29, 2017 7:07 am

### Re: Use of Rydberg formula

You can always derive the Rydberg Equation from E= hR/(n1)^2 - hR/(n2)^2 by dividing each side of the equation by h because the formula for frequency is E= hv.