Question

http://prntscr.com/d1tjqc I do not understand what I am doing wrong. Maybe I put something incorrectly in n1?

Answer 1

Balmer series formula given:\[v=(3.2881\times10^{15}s^{-1})\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)\]

Answer 2

that's the right formula, but i get a very different answer. i suspect your plugging in numbers is off

Answer 3

since \(n_2\) is 5, \((n_2)^2\) is 25, right?

Answer 4

It is? I used the Ryman one because things didn't seem to completely click.\[\frac{1}{\lambda}=R_{\infty}\left(\frac{1}{(n_{1})^{2}}-\frac{1}{(n_{2})^{2}}\right)\]

Answer 5

So I should revert back to Balmer series to solve this?\[\color{red}{n=5}\]\[v=(3.2881\times10^{15}s^{-1})\left(\frac{1}{2^{2}}-\frac{1}{\color{red}{5}^{2}}\right)\]

Answer 6

using the balmer formula gets you the \(frequency\), but then you use the equation for the speed of light \[c = \lambda * \nu\] to find the wavelength

Answer 7

that's what i did, and i got a very reasonable answer

Answer 8

getting a \(negative\) number for your wavelength should have been a big clue that something was wrong

Answer 9

Hmm...
I see what you mean, but is the Ryman series directly applicable in this scenario (for future reference)?

Answer 10

it is, but you get the \(inverse\) of wavelength, not the wavelength itself.

Answer 11

I did solve for \(\lambda\) by dividing 1 by the rest of the equation, if that is what you mean...

Answer 12

I think my main issue, though, is putting \(\large{n_{1}=8}\).

Answer 13

that might do it

Answer 14

\(n_2\) must always be larger than \(n_1\), so when you take the reciprocal of their squares, you still get a positive value

Answer 15

Hmm... I think for now I'll stick to your explanation of the Balmer series. I confused myself with the Ryman one haha

Answer 16

Okay... so \[v=(3.2881\times10^{15}s^{-1})(\frac{1}{2^{2}}-\frac{1}{5^{2}})=(3.2881\times10^{15})(\frac{1}{4}-\frac{1}{25})\]\[\approx6.90501\times10^{14}\]

Answer 17

Urgh, I got one digit off and it was incorrect... oh well

Answer 18

(I solved for \(\lambda\) already.)