If the wavelength of first member of Balmer series of hydrogen spectrum is 6564 A0, the wavelength of second member of Balmer series will be:

1λ∝1n12−1n22∴λ1λ2=122−142122−132⇒6564λ2=2720; ∴λ2=4862A∘

If the wavelength of first member of Balmer series of hydrogen spectrum is 6564 A0, the wavelength of second member of Balmer series will be:

1λ∝1n12−1n22∴λ1λ2=122−142122−132⇒6564λ2=2720; ∴λ2=4862A∘

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If the wavelength of first member of Balmer series of hydrogen spectrum is 6564 A⁰, the wavelength of second member of Balmer series will be:

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6050 A⁰

1215 A⁰

data given is insufficient to calculate the value

4862 A⁰

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{λ} \propto (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ ∴ \frac{λ_{1}}{λ_{2}} = \frac{\frac{1}{2^{2}} - \frac{1}{4^{2}}}{\frac{1}{2^{2}} - \frac{1}{3^{2}}} \Rightarrow \frac{6564}{λ_{2}} = \frac{27}{20}; ∴ λ_{2} = 4862 A^{\circ} \end{array}$