ionized lithium

logistic_guy

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Calculate the ionization energy of doubly ionized lithium, \(\displaystyle \text{Li}^{2+}\), which has \(\displaystyle Z = 3\).
 
Calculate the ionization energy of doubly ionized lithium, \(\displaystyle \text{Li}^{2+}\), which has \(\displaystyle Z = 3\).
Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem
 
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
Thank you Sir khan.

We start with the electron energy formula.

\(\displaystyle E_n = -\frac{Z^2(13.6 \ \text{eV})}{n^2}\)

The goal is to remove one electron from the nucleus completely, then:

The ionization energy \(\displaystyle = E_{\infty} - E_1 = 0 - \left(-\frac{3^2(13.6 \ \text{eV})}{1^2}\right) = 122 \ \text{eV}\)
 
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