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

Question: The 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 m ^-1

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

l  = 3645 [ n ² / ( 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  l (wavelength) to be a function of n as given above –such that we have:

3645 l ^-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 dl  = 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  ³  »  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.