Atomic hydrogen produces well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series fits the Rydberg equation with its own particular n₁ value. Calculate the value of n₁ (by trial and error if necessary) that would produce a series of lines in which:

(a) The highest energy line has a wavelength of 3282 nm.

(b) The lowest energy line has a wavelength of 7460 nm.

Answer: a) The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.

b) The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.

Explanation:

The formula that relates wavelength of emissions to Rydberg's constant and the n₁ values is

(1/λ) = R ((1 / (n₁^2)) - (1 / (n2^2))

Where λ = wavelength, R = (10.972 * 10^6) / m, n2 = ∞ (since they're emitted out of the atom already)

a) n₁ = ?

λ = 3282 nm = (3.282 * 10^-6) m

(1 / (3.282 * 10^-6)) = (10.972 * 10^6) ((1 / (n₁^2) (since 1/∞ = 0)

n₁^2 = (3.282 * 10^-6) * (10.972 * 10^6) = 36

n₁ = 6.

The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.

b) n₁ = ?

λ = 7460 nm = (7.46 * 10^-6) m

(1 / (7.46 * 10^-6)) = (10.972 * 10^6) ((1 / (n₁^2)) - (1 / (n2^2)) for lowest energy line, n2 = n₁ + 1

(n₁^2) ((n₁+1) ^2)) / (2n₁+1) = (7.46 * 10^-6) * (10.972 * 10^6) = 81.85

(n₁^2) ((n₁+1) ^2)) / (2n₁+1) = 81.85

Solving the quadratic eqn,

n₁ = 5.

The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.

QED!