QUANTUM PHYSICS FUNDAMENTALS - Chapter 3, Exercise 1 Solution
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Hydrogen Spectral Line: n=4 to n=2

PROBLEM
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Using the Rydberg formula, calculate the wavelength of light emitted
when a hydrogen electron falls from n = 4 to n = 2.

SOLUTION
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Using the real Rydberg formula:

    1/lambda = R_H (1/n1^2 - 1/n2^2)

Substitute the given values (n1 = 2, the lower final level; n2 = 4,
the higher starting level):

    R_H = 1.097 x 10^7 m^-1

    1/lambda = (1.097 x 10^7) x (1/2^2 - 1/4^2)
    1/lambda = (1.097 x 10^7) x (0.25 - 0.0625)
    1/lambda = (1.097 x 10^7) x 0.1875
    1/lambda = 2.057 x 10^6 m^-1

    lambda = 1 / (2.057 x 10^6)
    lambda = 4.86 x 10^-7 m
    lambda = 486 nm

ANSWER: The wavelength is approximately 486 nm.

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WHY THIS WORKS AS AN ANSWER

This result, 486 nm, is a real, verified match to one of the four
actual observed wavelengths in hydrogen's visible spectrum listed in
the chapter itself (410, 434, 486, and 656 nm) - confirming that the
Rydberg formula, and by extension Bohr's own theoretical model built
to explain it, correctly predicts real, measurable spectral lines.
The n=4-to-n=2 transition specifically produces this particular blue-
green line, distinct from the chapter's own worked n=3-to-n=2 example
(656 nm, red) - each specific pair of energy levels in the Balmer
series produces its own distinct, real, predictable wavelength.