If the wavelength of the first line of the Balmer series in the hydrogen spectrum is $λ$ , then the wavelength of the first line of the Lyman series is

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$(5 / 27) λ$

$(27 / 32) λ$

$(32 / 27) λ$

$(27 / 5) λ$

answer is B.

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Detailed Solution

For first line of Balmer series

$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}})$

$\Rightarrow R = \frac{36}{5 λ}$

$∴$ wavelength of the first line, $λ_{L}$ of the Lyman series is given by

$\frac{1}{λ_{L}} = R (1 - \frac{1}{2^{2}}) = \frac{36}{5 λ} \times \frac{3}{4} = \frac{27}{5 λ}$

$\Rightarrow λ_{L} = \frac{5 λ}{27}$

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