## 1.3 Atomic Spectra- Rydberg's equation

$c=\lambda v$

arina_m 1A
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am

### 1.3 Atomic Spectra- Rydberg's equation

Within this section(1.3 Atomic Spectra), we are using Rydberg's equation v= R(1/(n1)^2 - 1/(n2)^2) to find a certain frequency but I'm not sure I understand what wave's frequency it is that we are trying to calculate? And how exactly the components of the formula contribute to finding this frequency?

Chem_Mod
Posts: 18901
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 734 times

### Re: 1.3 Atomic Spectra- Rydberg's equation

Rydberg formula is used to calculate the frequency of light that is emitted as the hydrogen atom's electron falls from a higher excited state to a lower energetic state. It can be used vice-versa to figure out the frequency of light needed to excite the electron from a lower to a higher energy state in a hydrogen atom.

Nicole Shak 1L
Posts: 35
Joined: Wed Nov 22, 2017 3:03 am

### Re: 1.3 Atomic Spectra- Rydberg's equation

In class we found the change in energy, and used this to find the frequency and could subsequently find the wavelength. In the solutions manual they sue Rydberg's equation v= R(1/(n1)^2 - 1/(n2)^2), so I was also wondering if we should use this equation for problems since it's shorter or stick to the method shown in class?

Beverly Shih 1K
Posts: 34
Joined: Wed Nov 15, 2017 3:01 am
Been upvoted: 1 time

### Re: 1.3 Atomic Spectra- Rydberg's equation

Rydberg's equation is derived from the equation we used in class (E = -hR / n^2) with the main difference that Rydberg's equation has two n values, n1 and n2, so it can be used as a shortcut when you calculate the energy needed or released when transitioning between two energy levels. E = -hR / n^2 accomplishes the same task, but it might take more steps. But when you are only finding the current energy level of an electron, you should use the equation we used in class.