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Two years back my friend told me a simple formula for calculating the number of spectral lines. But, now I'm a bit confused about it number of lines is =$ \frac{2(n-1)}{2}$ is this right or is there any error in it?

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    $\begingroup$ minus infinity for the name $\endgroup$ – Jimmy360 Apr 9 '15 at 4:01
  • $\begingroup$ sorry! however can you help me out with the formula, like if hydrogen atom is excited to 4th level, it can either go from 4 to 3, 2, 1 or 3 to 2,1 or 2,1 which then adds up to 6 lines(lights that's emitted), $\endgroup$ – curiouslearner Apr 9 '15 at 4:14
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You can derive it simply by noting that each level can have $n-1$ transitions,so we have $n-1+n-2+...+1=n(n-1)/2$

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Proof time:

Each level can have $n-1$ transitions

This gives us $S = n-1 + n-2 + n-3+...+1$

Lets take $S$ and do this:

$S = n-1 + n-2 + n-3+...+1$

$+S = 1 + 2 + 3 +...+ n-1$

$= n(n-1)/2$

(because there are as we can tell from $S = 1 + 2 + 3 +...+ n-1$, there are $n-1$ elements in the series)

QED

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Let us not ignore the fact that there will be some overlaps too. For example, when there's a jump from 10 to 22 and a jump from 11 to 55, the spectral lines will overlap because 1/(10*10)-1/(22*22)=1/(11*11)-1/(55*55). Hence the actual number of spectral lines will be less than or equal to n*(n-1)*(1/2).

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  • $\begingroup$ This answer seems to be wrong. $\endgroup$ – Gonenc Jul 11 '15 at 16:46
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If a electron in hydrogen jumps from $n_1$ to $n_2$ then number of spectral lines is given by formula: $$\frac{(n_1-n_2)(n_1-n_2+1)}{2}$$

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    $\begingroup$ You might want to add an explanation of how that was arrived, simply stating the equation isn't particularly useful. $\endgroup$ – Kyle Kanos Jan 13 '16 at 15:07

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