COMEDK · Physics · 28. Atomic Physics
The ground state energy of hydrogen atom is \(-13.6 \mathrm{~eV}\). If the electron jumps from the \(3^{\text {rd }}\) excited state to the ground state then the energy of the radiation emitted will be:
- A 12.75 eV
- B 12.75 MeV
- C 1.275 MeV
- D 12.75 J
Answer & Solution
Correct Answer
(A) 12.75 eV
Step-by-step Solution
Detailed explanation
The energy of an electron in the \(n^{th}\) orbit of a hydrogen atom is given by \(E_n = -\dfrac{13.6}{n^2} \text{ eV}\). The ground state corresponds to \(n = 1\). The \(3^{rd}\) excited state corresponds to \(n = 1 + 3 = 4\). The energy of the electron in the ground state is…
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