Question

The frequency of radiation emitted when electron falls from n=4 to n=1 in a hydrogen atom will be (given ionization energy of H = `2.18 xx 10^(-18) "J/atom"` and `h = 6.626 xx 10^(-34) "J"cdot"s"`)?

Answer 1

`nu_(4->1) = 3.084 xx 10^(15) "s"^(-1)`

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Electronic transitions in the hydrogen atom are given by the Rydberg equation:

`bb(DeltaE = -R_H(1/n_f^2 - 1/n_i^2))`

where:

• `R_H = 2.18 xx 10^(-18) "J/H atom"` is the Rydberg constant. Yes, it is the ionization energy of hydrogen atom as well!
• `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant.
• `n_k` is the principal quantum number for the `k`th state. Thus, `n_f` indicates the destination energy level, and `n_i` indicates the initial energy level.

So, `n = 4 -> 1` implies `n_f = 1` and `n_i = 4`. This means...

`DeltaE = -2.18 xx 10^(-18) "J/H atom" xx (1/1^2 - 1/4^2)`

which should be negative with respect to the system! (why?)

...and we do indeed get a negative number...

`DeltaE = -2.044 xx 10^(-18) "J/H atom"`

The energy released from the `"H"` atom due to the relaxation process contains photons, which each have energy

`|DeltaE| = E_"photon" = hnu`

where `nu` is the frequency in `"s"^(-1)`.

Thus, the energy per photon is:

`E_"photon" = +2.044 xx 10^(-18) "J/H atom"`

`= hnu`

and the frequency is therefore:

`color(blue)(nu) = E_"photon"/h = (+2.044 xx 10^(-18) cancel"J"/cancel"H atom" xx cancel"1 H atom"/"photon")/(6.626 xx 10^(-34) cancel"J"cdot"s")`

`= color(blue)ul(3.084 xx 10^(15) "s"^(-1))`