JEE Mains · Physics · STD 12 - 12. atoms
The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is \(915\)\(A\). The longest wavelength of spectral lines in the Balmer series will be _______ \(A\).
- A \(6587\)
- B \(6588\)
- C \(6590\)
- D \(6596\)
Answer & Solution
Correct Answer
(A) \(6587\)
Step-by-step Solution
Detailed explanation
Lyman Series Shortest, \(\frac{\mathrm{hc}}{\lambda}=-13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\) \(\lambda \downarrow \mathrm{E} \uparrow ; \frac{\mathrm{hc}}{\lambda_0}=-13.6(1)\) Balmer Series: \(\mathrm{n}=3\) \(\mathrm{n}=2\)…
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