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Q.

In hydrogen like atom (Z = 11), n^th line of Lyman series has wavelength $λ$ equal to the de-Broglie’s wavelength of electron in the level from which it originated. The value of n is

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a

n = 2

b

n = 10

c

n = 8

d

n = 25

answer is D.

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Detailed Solution

$\frac{1}{\lambda } = R{Z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)\$
$= R{\left( {11} \right)^2}\left( {\frac{1}{1} - \frac{1}{{{n^2}}}} \right)\$
$\lambda = \frac{h}{{mv}}\,(de - broglie\,equation)\$
$\lambda = \frac{{2\pi r}}{n} = \frac{{\pi \left( {0.529 \times {{10}^{ - 10}}} \right){n^2}}}{{n\left( {11} \right)}}\$ (bonr’s quantization for angular momentum )
$\Rightarrow \frac{1}{\lambda } = \frac{{11}}{{2\pi \left( {0.529 \times {{10}^{ - 10}}} \right)n}}\$
⇒ n = 25 (on solving)

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