ExamBro
ExamBro
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.
Verified Solution

Answer & Solution

Correct Answer

(D) 3

Step-by-step Solution

Detailed explanation

\(\Delta \mathrm{E}=13.6 \mathrm{z}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\)
\((13.6) \mathrm{z}^2\left[\frac{1}{1}-\frac{1}{9}\right]=108.8\)
\(\frac{(13.6)(8)}{9}\left(z^2\right)=108.8\)
\(\mathrm{z}=3\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app