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Q.

The wavelength of the first member of the Balmer series in hydrogen spectrum is x A⁰. Then the wave length (in A⁰) of the first member of Lyman series in the same spectrum is

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a

5/27 x

b

4/3 x

c

5/36 x

d

27/5 x

answer is A.

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Detailed Solution

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$\large \bar v = \frac{1}{\lambda } = {R_H}{Z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)\$

H- atom, Z = 1

Balmeseries n₁ = 2

first member, n₂ = n₁ + 1

= 2 + 1 = 3

$\large \bar v = \frac{1}{\lambda } = {R_H}{1^2}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\$

$\large \bar v = \frac{1}{\lambda } = \frac{{5{R_H}}}{{36}}\$

$\large \lambda = \frac{{36}}{{5{R_H}}} = x{A^0}\$ (given)

$\large {R_H} = \frac{{36}}{{5x}}....(1)\$

H-atom, Z = 1

Lyman series, n₁ = 1

First member = n₁ + 1

= 1 + 1 = 2

$\large \bar v = \frac{1}{\lambda } = {R_H}{1^2}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)\$

$\large = \frac{{3{R_H}}}{4}\$

$\large \lambda = \frac{4}{{3{R_H}}}\$

Substituting the value of R_H from (1) equation

$\large \lambda = \frac{4}{{3\left( {\frac{{36}}{{5x}}} \right)}}\$

= 5x/27

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