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Find the potential energy of electron in first excited state if in the n^th state, energy of electron in hydrogen atom is $- \frac{13.6 e V}{n^{2}}$

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−27.2eV

−3.4eV

−3.2eV

−6.8eV

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detailed solution

Correct option is D

TE=−13.6eV(2)2=−3.4eVPE=−6.8eV

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Find the potential energy of electron in first excited state if in the nth state, energy of electron in hydrogen atom is -13.6 eVn2

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Ready to Test Your Skills?

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Find the potential energy of electron in first excited state if in the n^th state, energy of electron in hydrogen atom is $- \frac{13.6 e V}{n^{2}}$