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If the wave number of 1^st line of Balmer series of H-atom is ‘x’ then

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a

the wave length of 2^nd line of lyman series of H-atom is $\frac{5}{32 x}$

b

wavenumber of 1st line of lyman series of He⁺ ion will be $\frac{36 x}{5}$

c

the wave length of 2^nd line of lyman series of H-atom is $\frac{32 x}{5}$

d

wavenumber of 1st line of lyman series of He⁺ ion will be $\frac{108 x}{5}$

answer is A, C.

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Detailed Solution

$\begin{array}{l} \bar{v} = {RZ}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ x = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R}{36} \\ {\bar{v}}_{1} = R \times 2^{2} (1 - \frac{1}{2^{2}}) = 3 R = \frac{36}{5} x \times 3 = \frac{108 x}{5} \end{array}$

For H atom 2^nd lyman

$\frac{1}{λ_{,}} = R (1 - \frac{1}{9}) = \frac{8}{9} \times \frac{36 x}{5} = \frac{32 x}{5} \Rightarrow λ_{2} = \frac{5}{32 x}$

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