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.
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.