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

Jun 22, 2017

About $\text{1.89 eV}$ is released for the absorption, or absorbed for the emission process between $n = 3$ and $n = 2$. That means it takes about $1.89$ $\text{eV}$ 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).

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:

$\Delta E = - \text{13.6 eV} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

where:

• $\text{13.6 eV}$ 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:

$\Delta E = \textcolor{b l u e}{{E}_{3} - {E}_{2}} = - \text{13.6 eV} \left(\frac{1}{3} ^ 2 - \frac{1}{2} ^ 2\right)$

$=$ $\textcolor{b l u e}{\text{1.89 eV}}$