Question

Electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting lightofl = 93.8nm.find the principal level to which the electron relaxed.

Answer 1

The electron in the n = 6level of the hydrogen atom, relaxing and emitting light of
`\(93.8 \ \text{nm}\)`, transitions to the n = 3 principal level.

Step-by-step explanation:

When an electron transitions from a higher energy level n_i to a lower energy level n_f in a hydrogen atom, it emits light with a wavelength given by the Rydberg formula:

`\[ (1)/(\lambda) = R_H \left((1)/(n_f^2) - (1)/(n_i^2)\right), \]`

where R_H is the Rydberg constant for hydrogen
`(\(1.097 * 10^7 \ \text{m}^(-1)\)), \(\lambda\)` is the wavelength of the emitted light, and n_i and n_f are the principal quantum numbers of the initial and final energy levels, respectively.

In this case, the electron transitions from
`\(n_i = 6\) to \(n_f\)`, and the wavelength of the emitted light is given as
`\(93.8 \ \text{nm}\)`. By rearranging the formula, we can solve for
`\(n_f\)`:

`\[ (1)/(93.8 * 10^(-9)) = 1.097 * 10^7 \left((1)/(n_f^2) - (1)/(6^2)\right). \]`

Solving this equation yields
`\(n_f = 3\),` indicating that the electron relaxes to the
`\(n = 3\)` principal level. Understanding these quantum transitions is fundamental to explaining the spectral lines observed in hydrogen atoms.