JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2020
- Medium
- Numerical
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
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Answer & Solution
Correct Answer
(C) \(486\)
Step-by-step Solution
Detailed explanation
For Balmer series.
\(\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)\)
\(\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)}\)
\(\frac{\lambda_{2}}{6561}=\frac{5 / 36}{3 / 16}\)
\(\lambda_{2}=\frac{20}{27} \times 6561\)
\(\lambda_{2}=4860 \mathring{A}\)
\(=486 \mathrm{nm}\)
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