Last line of Lyman series for H-atom wavelength $λ_{1} Å, 2 n d$ line of Balmer series has wavelength $λ_{2} Å$ then:

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$\frac{16}{λ_{1}} = \frac{9}{λ_{2}}$

$\frac{16}{λ_{1}} = \frac{3}{λ_{2}}$

$\frac{16}{λ_{2}} = \frac{3}{λ_{1}}$

$\frac{4}{λ_{1}} = \frac{1}{λ_{2}}$

answer is B.

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Detailed Solution

If pura question is,

$\begin{array}{l} \frac{1}{λ_{1}} = R_{H} [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}]; \frac{1}{λ_{2}} = R_{H} [\frac{1}{2^{2}} - \frac{1}{4^{2}}] \\ \Rightarrow \frac{1}{λ_{2}} = \frac{1}{λ_{1}} \times [\frac{3}{16}] \end{array}$

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