## Thursday, November 8, 2018

### Selected Questions-Answers Form All Experts Astronomy Forum (The Balmer Series Of Spectral Lines)

QuestionThe lines of the Balmer series crowd close together as higher series members are considered. So if each line were exactly one A (1 Angstrom) wide, how many Balmer lines would be individually visible without overlapping other lines? (N.B. This is not a homework question!)

Answer:  Thanks for clarifying this is not a homework problem since it did look suspicious.

Anyway,  the Balmer lines are defined according to:

1/ l = R (1/ 4 – 1/n 2)

where l defines the wavelength, n is an energy level > 2 and R is the Rydberg constant:

R = 1.097 x 10 7 -1

From this one obtains (with appropriate algebra, change of units):

l  = 3645 [ / ( n 2 – 4)]

which yields all wavelengths in Angstroms, unlike the original expression which yields wavelength in meters. (Bear in mind here the need to pay attention to the units, since 1 A = 10 -10 m )

Overlapping of lines must begin for that ‘n’ for which we have the condition:

l(n) – l(n + 1) = 1 A

Now, consider  (wavelength) to be a function of n as given above –such that we have:

3645 -1 =  1 – 4n -2

The next (and last) step in this solution, is that one makes use of differentials (after differentiating) such that:

-3645  l -2 d = 8n -3 dn

From here, it is straightforward. Simply solve for  n  -3  on one side, with the derivative:    (-dn/dl) on the other.

Then:

n   »  8 l 2 / 3645  [ (-dn/dl) ]   ,

Which specifies the value for n for which dn = 1 and dl = -1

It's evident that this n is considerably greater than 1, so  l must be very close to 3645A

or n »  31 => (n + 1) =   (30 + 1) =  1 A

That is, about 30 Balmer lines are visible.