Question

when an excited electron in a hydrogen atom falls from n=6 to n= 2, a photon of violet light is emitted. if an excited electron in an He+ ion falls from n=5, which energy level must it fall to (n1) for violet light of a similar wavelength to be emitted

Answer 1

The energy level the electron must fall to in order to emit violet light of a similar wavelength is n1 = 4.

The energy of a photon emitted during a transition in a hydrogen-like atom is given by the Rydberg formula:

`\[ (1)/(\lambda) = R \left((1)/(n_1^2) - (1)/(n_2^2)\right) \]`

In the given scenario, for the hydrogen atom
`(\(H\))`, the electron transitions from
`\(n=6\)` to
`\(n=2\)` emitting violet light.

For a similar wavelength in an
`\(He^+\)` ion, which has one less electron than hydrogen, the transition would be from
`\(n=5\)` to
`\(n_1\)`.

By substituting these values into the Rydberg formula, we find that
`\(n1 = 4\)` for violet light to be emitted.