Answer to Question #118913 in Optics for Shanil

Question

You have a scriptbase text Li end text end scriptbase presubscript 3 presuperscript 7 atom (3 protons) which is doubly ionized such that it has just one remaining electron. (scriptbase text Li end text to the power of 2 plus end exponent end scriptbase presubscript 3 presuperscript 7)

How much energy, in electron volts, does it take to remove the remaining electron if it starts in the n = 1 orbital? (to 2 s.f and in eV)

Answer 1

The ionization energy required is

$$E=13.6\frac{Z^2}{n^2}=13.6\frac{3^2}{1^2}=122\text{ eV}.$$