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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) \(486\)

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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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