JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2024
- Hard
- Numerical
Radius of a certain orbit of hydrogen atom is \(8.48\mathring A \). If energy of electron in this orbit is \(\mathrm{E} / \mathrm{x}\), then \(\mathrm{x}=\) _______. (Given \(a_0=0.529 \mathring A, E=\) energy of electron in ground state)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
We know
\(\mathrm{r}=0.529 \frac{\mathrm{n}^2}{\mathrm{Z}} \Rightarrow 8.48=0.529 \frac{\mathrm{n}^2}{1}\)
\(\mathrm{n}^2=16 \Rightarrow \mathrm{n}=4\)
We know
\(E \propto \frac{1}{n^2}\)
\(E_{n^{\text {dh }}}=\frac{E}{16}\)
\(x=16\)
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