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

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

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Answer 1 · 2017-06-22T04:18:22

About $1.89 eV$ $"1.89 eV"$ is released for the absorption, or absorbed for the emission process between $n = 3$ $n = 3$ and $n = 2$ $n = 2$ . That means it takes about $1.89$ $1.89$ $eV$ $"eV"$ 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 $n = 3$ $n = 3$ to $n = 2$ $n = 2$ 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:

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

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

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

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

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