JEE Mains · Physics · STD 12 - 12. atoms
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)
- A \(14\)
- B \(15\)
- C \(16\)
- D \(20\)
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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