Question

What is the ionization potential for an electron going from `n = 2` to `n = 3` in `"eV"` for hydrogen atom?

Answer 1

About `"1.89 eV"` is released for the absorption, or absorbed for the emission process between `n = 3` and `n = 2`. That means it takes about `1.89` `"eV"` of work to accelerate one electron through a one potential difference of one volt.

To be accurate, I don't think that's an ionization potential... that's really the energy used to excite the atom (ionization potential is for removing electron(s) altogether).

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I think you're asking about the change in energy going from `n = 3` to `n = 2` in the hydrogen atom? That can be found from the Rydberg equation:

`DeltaE = -"13.6 eV" (1/n_f^2 - 1/n_i^2)`

where:

• `"13.6 eV"` is the Rydberg constant in electron volts (or the negative of the hydrogen atom ground state energy).
• `n_i` is the initial state's principal quantum number `n`.
• `n_f` is the final state's principal quantum number `n`.

So, if you wanted the final state to be `E_3`, then:

`DeltaE = color(blue)(E_3 - E_2) = -"13.6 eV" (1/3^2 - 1/2^2)`

`=` `color(blue)("1.89 eV")`