WBJEE · Physics · Atomic Physics
The ionization energy of the hydrogen atom is 13.6 eV. The potential energy of the electron in \(n=2\) state of hydrogen atom is
- A \(+3.4 \mathrm{eV}\)
- B \(-3.4 \mathrm{eV}\)
- C \(+6.8 \mathrm{eV}\)
- D \(-6.8 \mathrm{eV}\)
Answer & Solution
Correct Answer
(D) \(-6.8 \mathrm{eV}\)
Step-by-step Solution
Detailed explanation
Given, \(E_{n}=13.6 \mathrm{eV}\) Energy of an electron in \(n^{\text {th }}\) state \(E_{n}=\frac{-13.6 z^{2} \mathrm{eV}}{n^{2}}\) \(\therefore\) Energy of an electron in \(n=2\) state \(E_{2}=\frac{18-6 z^{2}}{(2)^{2}}=-34 e V\) So.…
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