Question

Question #7e7e4

Answer 1

The frequency of a photon emitted or absorbed in that electronic transition is `6.9 times 10^14 color(white) {i} Hz`.

Explanation:

The wavelength of the photon involved is given by the Rydberg's formula:

`1/lambda = R_H (1/n_1^2 - 1/n_2^2)`,

where `R_H` is the known Rydberg's constant, `R_H = 1.097 times 10^7` `m^{- 1}` and the numbers `n_1` and `n_2` are the initial and final first quantic numbers of energy levels of the electron.

Then we have:

`1/lambda = 1.097 times 10^7` `m^{- 1} cdot (1/2^2 - 1/5^2) = 2.304 times 10^6` `m^{- 1}`.

The radiation frequency can be obtained from the equation of the propagation velocity in the vacuum:

`c = lambda cdot nu` `rArr` `nu = c/lambda = c cdot 1/lambda`

Then:

`nu = 3 times 10^8` `m cdot s^{- 1} cdot 2.304 times 10^6` `m^{- 1} = 6.9 times 10^14` `Hz`.