KCET · Physics · Atomic Physics
Hydrogen atom from excited state comes to the ground stage by emitting a photon of wavelength \(\lambda\). If \(R\) is the Rydberg constant, the principal quantum number \(n\) of the excited state is
- A \(\sqrt{\frac{\lambda R}{\lambda R-1}}\)
- B \(\sqrt{\frac{\lambda}{\lambda R-1}}\)
- C \(\sqrt{\frac{\lambda \mathrm{R}^{2}}{\lambda \mathrm{R}-1}}\)
- D \(\sqrt{\frac{\lambda R}{\lambda-1}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{\lambda R}{\lambda R-1}}\)
Step-by-step Solution
Detailed explanation
Here, \(\mathrm{n}_{\mathrm{f}}=1, \mathrm{n}_{\mathrm{i}}=\mathrm{n}\)
\(\therefore \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{\mathrm{n}^{2}}\right) \Rightarrow \frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{\mathrm{n}^{2}}\right) \quad \text{...(i)}\)
or \(\frac{1}{\lambda R}=1-\frac{1}{n^{2}}\) or \(\frac{1}{n^{2}}=1-\frac{1}{\lambda R}\)
or \(\quad \mathrm{n}=\sqrt{\frac{\lambda R}{\lambda R-1}}\)
\(\therefore \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{\mathrm{n}^{2}}\right) \Rightarrow \frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{\mathrm{n}^{2}}\right) \quad \text{...(i)}\)
or \(\frac{1}{\lambda R}=1-\frac{1}{n^{2}}\) or \(\frac{1}{n^{2}}=1-\frac{1}{\lambda R}\)
or \(\quad \mathrm{n}=\sqrt{\frac{\lambda R}{\lambda R-1}}\)
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