Atomic numbers of two elements are 31 and 41. Find the ratio of the wavelength of the ${K_\alpha }\$ are

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4 : 3

9 : 16

16 : 9

16 : 27

answer is D.

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Detailed Solution

$\frac{1}{\lambda } = R{(z - b)^2}\left( {\frac{1}{{{n_1}^2}} - \frac{1}{{{n_2}^2}}} \right)\$

$\therefore \frac{1}{\lambda }\alpha {(z - b)^2},for\,\,K - series,b = 1\$

$\therefore \frac{1}{\lambda }\alpha {(z - 1)^2} \Rightarrow\,\, \therefore \frac{{{\lambda _1}}}{{{\lambda _2}}} = {\left( {\frac{{{z_2} - 1}}{{{z_1} - 1}}} \right)^2} = {\left( {\frac{{41 - 1}}{{31 - 1}}} \right)^2} = 16:9\$

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