Question

# The electron in a hydrogen atom falls from an excited energy level to the ground state...

The electron in a hydrogen atom falls from an excited energy level to the ground state in two steps, causing the emission of photons with wavelengths of 656.5 nm and 121.6 nm (So the in the first step the 656.5 nm photon is emitted and in the second step the 121.6 nm photon is emitted). What is the principal quantum number (ni) of the initial excited energy level from which the electron falls?

According to the Rydberg equation:

1 / λ = RH * (1 / ni ^ 2 - 1 / nf ^ 2)

RH: Rydberg constant for hydrogen. 10967758.341 m-1

For the first step: From 1 to unknown level, with wavelength 121.6 nm = 1.216x10 ^ -7 m, clear and replace:

1 / nf ^ 2 = 1/1 ^ 2 - 1 / λ * RH = 1/1 - 1 / (1.216x10 ^ -7 * 10967758.341) = 0.2502

Clearing:

nf = √1 / 0.2502 = 2

our energy level was 2.

Second step, from level 2 to the final unknown level, with a wavelength of 656.5 nm = 6.565 x10 ^ -7 m:

1 / nf ^ 2 = 1/2 ^ 2 - 1 / (6.565x10 ^ -7 * 10967758.341) = 0.1111

nf = √1 / 0.1111 = 3

final energy level was 3.

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