Chemistry, asked by Pradyumna1094, 11 months ago

Potential energu is 27.2 in second orbit of helium , then calculate double of total energy in first excited state of hydrogen atom

Answers

Answered by Anonymous

1

Explanation:

Energy(Helium) = −13.6×4n2

PE=−2KE.

TE=PE2=−27.22×2=−6.8 eV

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