Question

What is the wave number of the hydrogen atom excitation from `n = 1` to `n = 3`?

Answer 1

About `"97533 cm"^(-1)`.

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I assume you mean, what is the frequency in `"cm"^(-1)` (i.e. wavenumbers) of the electronic excitation `n = 1 -> 3` in the hydrogen atom...

We consult the Rydberg equation in `"eV"` (because I am too lazy to remember the Rydberg constant in `"m"^(-1)` or what have you):

`DeltaE = -"13.606 eV"(1/n_f^2 - 1/n_i^2)`

where:

• `n_k` is the principal quantum number for the `k`th state in the hydrogen atom, i.e. initial `i` or final `f`.
• `-"13.606 eV"` is the approx. ground state energy of hydrogen atom.

Thus, the change in energy for this excitation (which is positive!) is:

`DeltaE = -"13.606 eV"(1/3^2 - 1/1^2)`

`=` `"12.094 eV"`

And now it is a matter of unit conversion into the equivalent frequency in `"cm"^(-1)`. The following constants will be useful:

• `h = 6.626 xx 10^(-34) "J"cdot"s"`, Planck's constant.
• `c = 2.998 xx 10^(10) "cm/s"`, the speed of light.
• `e ~= 1.602 xx 10^(-19) "J/eV"`, numerically equal to the elementary charge.

`color(blue)(Deltanu) = 12.094 cancel"eV" xx (1.602 xx 10^(-19) cancel"J")/(cancel"1 eV")`

`xx 1/((2.998 xx 10^(10) "cm/"cancel"s")(6.626 xx 10^(-34) cancel"J" cdotcancel"s"))`

`~~ color(blue)("97533 cm"^(-1))`