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Q.

The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 A⁰. The wavelength of the first line is

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a

$\frac{27}{20} \times 4861 \overset{0}{A}$

b

$\frac{20}{27} \times 4861 \overset{0}{A}$

c

$20 \times 4861 \overset{0}{A}$

d

$4861 \overset{0}{A}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{λ} = {RZ}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}); ∴ \frac{λ_{1}}{λ_{2}} = \frac{\frac{1}{2^{2}} - \frac{1}{4^{2}}}{\frac{1}{2^{2}} - \frac{1}{3^{2}}} \\ ∴ \frac{λ_{1}}{4861} = \frac{27}{20}; ∴ λ_{1} = \frac{27}{20} \times 4861 A^{0} \end{array}$

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The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 A0. The wavelength of the first line is