JEE Mains · Physics · STD 12 - 12. atoms
A hydrogen atom makes a transition from \(n = 2\) to \(n = 1\) and emits a photon. This photon strikes a doubly ionized lithium atom \((z = 3)\) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is
- A \(2\)
- B \(4\)
- C \(5\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
A hydrogen atom makes a transition from \(n=2\) to \(n=1\) Then wavelength \(=\operatorname{Rcz}^{2} \frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}=\operatorname{Rc}(1)^{2}\left[1-\frac{1}{4}\right]\) \(\lambda=\operatorname{Rc}\left[\frac{3}{4}\right]\) ..... \((1)\)…
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