AP EAMCET · PHYSICS · Atomic Physics
If the bonding energy of the electron in a hydrogen atom is 13.6 eV , then energy required to remove electron from first excited state of \(\mathrm{Li}^{2+}\) is
- A 122.4 eV
- B 3.4 eV
- C 13.6 eV
- D 30.6 eV
Answer & Solution
Correct Answer
(D) 30.6 eV
Step-by-step Solution
Detailed explanation
\(B . E=13.6 \mathrm{eV}=-E_1 \Rightarrow E_1=-13.6 \mathrm{eV}\) For \(\mathrm{Li}^{2+}, \mathrm{z}=3, \mathrm{n}_1=2, \mathrm{n}_2=\infty\) \(\therefore \quad \Delta \mathrm{E}=\mathrm{E}_{\infty}-\mathrm{E}_2=0-\left(-13.6 \times \frac{\mathrm{Z}^2}{\mathrm{~n}_1^2}\right)\)…
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