A) 3.4 eV
B) 6.8 eV
C) 13.6 eV
D) 1.7 eV
Correct Answer: A
Solution :
Key Idea: Total energy of electron in the orbit is equal to negative of its kinetic energy. The energy of hydrogen atom when the electron revolves in nth orbit is \[E=\frac{-13.6}{{{n}^{2}}}eV\] In the ground state; n = 1 \[\therefore E=\frac{-13.6}{{{1}^{2}}}=-13.6\,eV\] For \[n=2,\,E=\frac{-13.6}{{{2}^{2}}}=-3.4\,eV\] So, kinetic energy of electron in the first excited state (i.e., for n = 2) is \[K=-E\,=(-3.4)=3.4\,\,eV\]
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