JEE Mains · Physics · STD 12 - 12. atoms
A particular hydrogen like ion emits radiation of frequency \(2.92 \times 10^{15}\, {Hz}\) when it makes transition from \({n}=3\) to \({n}=1 .\) The frequency in \({Hz}\) of radiation emitted in transition from \({n}=2\) to \({n}=1\) will be : (in \(\,\times 10^{15}\))
- A \(0.44\)
- B \(6.57\)
- C \(4.38\)
- D \(2.46\)
Answer & Solution
Correct Answer
(D) \(2.46\)
Step-by-step Solution
Detailed explanation
\(\mathrm{nf}_{1}=\mathrm{k}\left(\frac{1}{1}-\frac{1}{3^{2}}\right)\) \(\mathrm{nf}_{2}=\mathrm{k}\left(1-\frac{1}{2^{2}}\right)\) \(\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{8 / 9}{3 / 4} \Rightarrow \mathrm{f}_{2}=2.46 \times 10^{15}\)
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