JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2025
- Medium
- MCQ
In a hydrogen like ion, the energy difference between the \(2^{\text {nd }}\) excitation energy state and ground is 108.8 eV. The atomic number of the ion is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{E}=13.6 \mathrm{z}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\)
\((13.6) \mathrm{z}^2\left[\frac{1}{1}-\frac{1}{9}\right]=108.8\)
\(\frac{(13.6)(8)}{9}\left(z^2\right)=108.8\)
\(\mathrm{z}=3\)
\((13.6) \mathrm{z}^2\left[\frac{1}{1}-\frac{1}{9}\right]=108.8\)
\(\frac{(13.6)(8)}{9}\left(z^2\right)=108.8\)
\(\mathrm{z}=3\)
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