Question

Estimate the energy required to promote two electrons to the lowest excited state of the He atom. What happens if this amount of energy is absorbed by the He atom in its ground state?

Answer 1

lowest excited state means n=2
for 1 electron,
E = 13.6*Z^2 (1/n1^2 - 1/n2^2)
= 13.6*(2)^2 (1/1^2 - 1/2^2)
= 13.6*4* (0.75)
= 40.8 eV

for 2 electron energy required = 81.6 eV
Answer: 81.6 eV

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If this energy if provided to 1 atom :
E nergy required to ionize 1 He atom =E = 13.^*z^2
= 13.6*4
=54.4 eV

So, 81.6 eV will be enougth to ionize an electron
So helium ion will be formed from the atom if that amount is absorbed by He atom