JEE Mains · Physics · STD 12 - 12. atoms
The first member of the Balmer series of hydrogen atom has a wavelength of \(6561\; \mathring A\). The wavelength of the second member of the Balmer series (in \(nm\)) is
- A \(256\)
- B \(540\)
- C \(486\)
- D \(626\)
Answer & Solution
Correct Answer
(C) \(486\)
Step-by-step Solution
Detailed explanation
For Balmer series. \(\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)\) \(\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)}\)…
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