NEET · Physics · STD 12 - 12. atoms
The total energy of an electron in the \(n^{t h}\) stationary orbit of the hydrogen atom can be obtained by
- A \(E_{n}=-13.6 \times n^{2} \;eV\)
- B \(E_{n}=\frac{13.6}{n^{2}} \;eV\)
- C \(E_{n}=-\frac{13.6}{n^{2}} \;eV\)
- D \(E_{n}=13.6 \times n^{2} \;eV\)
Answer & Solution
Correct Answer
(C) \(E_{n}=-\frac{13.6}{n^{2}} \;eV\)
Step-by-step Solution
Detailed explanation
Energy of electron in \(n^{\text {th }}\) orbit of hydrogen atom, \(E_{n}=-\frac{13.6}{n^{2}} e V\)
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