## Atomic Spectra

$c=\lambda v$

Tim Foster 2A
Posts: 73
Joined: Fri Sep 29, 2017 7:07 am

### Atomic Spectra

Using the formula
V=R(1/nsquared - 1/nsquared)
in atomic spectra problems, what does it mean when the initial n is greater than the final n (ex// n=1 transitioning to n=5) and you get a negative value for the frequency of the energy emitted?

Nancy Dinh 2J
Posts: 59
Joined: Fri Sep 29, 2017 7:07 am

### Re: Atomic Spectra

You won't get a negative value because the value for $n_{1}$ (initial) is always going to be lower than $n_{2}$ (final). When you put those values under a fraction, the previously larger value will become the smaller one, and vice versa.

Let's use your example of 1 and 5:
$v=R(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$

$v=R(\frac{1}{{1^{2}}-\frac{1}{{5}^{2}}})$
$v=R(1-\frac{1}{25})$
$v=R(\frac{24}{25})$

Hope this helps!