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How to calculate the number of spectral lines emitted or absorbed ?

Rydberg formula: 1/λ = RH(1/n2 – 1/m2), where λ = the wavelenght, RH = 10973731.55 cm-1 is the Rydberg constant for hydrogen and n, m are integers.For the Balmer series in the visible region
the formula is as follows: 1/λ = RH(1/4 – 1/m2), m = 3, 4, 5, ...

So you need two terms: from which line the emission or absorbtion took place and wavelength. Rest can be calculated.

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Number of spectral lines from a state n to ground state is :  [n(n - 1) ] / 2
 
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hiiii,,, 

suppose an electron falls from outer orbit n2 to inner orbit n1 or vise versa,,,,,,then no. of spectral lines emitted or absorbed= { ( n2 - n1) ( n2 - n1 + 1)} / 2

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@darpan : yr ye ..... { ( n2 - n1) ( n2 - (n1 + 1)} / 2 ye hoga ........... tumne opening bracket nahi lagaya  ....... 

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nhi yr opening bracket nhi ayega,....................    i am dammed sure....

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mr.darpan is totally rite...no opening bracket ..

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thankzzz Hitesh

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 no. of spectral lines emitted is equal to the photons absorbed

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I think Mr. PoET is correct !!

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Mee 2
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it depends
if u calculate maximum spectral lines of a single h-atom then it is (n-1)
and if u calculate maximum spectral lines of sample of h-atom then it is (n(n-1)/2)
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above given answers are helpful
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formula is :- n(n-1)/2
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n(n-1)/2
 
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(n2-n1)?(n2-n1+1)/2
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the ground state energy of hydrogen atom is -13.6 EV if an electron make a transition from an energy level -1.51 EV to -3.4 EV calculate the wavelength of the spectral light emitted and the name the series of hydrogen spectrum to which it belongs
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(N1-N2)(N1-N2+1)/2
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possible spectral lines=(n2-n1)(n2-n1-1)/2
total no. of spectral lines from n=n to n=1 = n(n-1)/2
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sorry...n1+1 
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n(n-1)/2
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n(n-1)/2
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If n1 be the initial shell no. and n2 be the final shell no., then no.of spectral lines is given by [(n2-n1)(n2-n1+1)]/2

This is valid when question is asked for finding total no.of spectral lines emitted due to transitions associated with all shells.

But if question is asked to find the no.of spectral lines associated with only a particular shell(n), then formula is [n(n-1)]/2
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No of spectral lines =(n2-n1) (n2-n1+1)/ 2
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To find no of spectral lines 2 formula use generally : 1 if electron is comming from any sub shell to the ground state use formula 👉 n(n-1)/2 2 if electron is comming on other than ground state use formula👉(n2-n1)(n2-n1+1)/2 .... u will get the answer definitely Thank q
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Please find this answer

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The formula is (n2-n1)(n2-n1+1)/2
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Regards

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n(n-1)/2
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Nicole pauly
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n?-n/2
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No. of spectral lines in ground state= n( n-1) Where 'n' is no. of orbit from where e is coming If e comes to excited state then no. of spectral lines =(N-n)(N-n+1) Where'N' is the orbit from where e is coming and'n' is orbit where e is coming
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And : emitted = n(n-1)/2
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Please find this answer

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Please find this answer

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8.8×1014 m/sec2
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n(n+1)by 2
.. n=no. of gaps present in spectral lines.. for example from fifth position... 5-1=4 so 4(4+1) by 2 =10
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Please find this answer

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formula is n(n+1)/2
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N(N-1)/2 WhereN is qiantum no energy level
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Please find this answer

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(n2-n1)(n2-n1+1)/2
(or)
n(n-1)/2
Regards
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You know what is answer
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Calculate last principles qontem number after you can calculate number of spectral lines
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The formula to calculate it is: n(n-1)/2
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Please find this answer

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Please find this answer

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ano ang
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-222from0
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1/lambda = Rh((1/(nf^2))-(1/(ni^2)))
 
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