__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.

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