The wavelength of L_{alpha} line in X-ray spe | vedclass

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Question

(B) According to Moseley's Law,the frequency $
u$ of a characteristic $X$-ray line is given by $\sqrt{
u} = a(Z - b)$,where $Z$ is the atomic number

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(B) According to Moseley's Law,the frequency $
u$ of a characteristic $X$-ray line is given by $\sqrt{
u} = a(Z - b)$,where $Z$ is the atomic number,$b$ is the screening constant,and $a$ is a constant.Since $
u = \frac{c}{\lambda}$,we have $\sqrt{\frac{c}{\lambda}} = a(Z - b)$,which implies $\frac{1}{\sqrt{\lambda}} \propto (Z - b)$.For the two elements,we can write the ratio: $\sqrt{\frac{\lambda_2}{\lambda_1}} = \frac{Z_1 - b}{Z_2 - b}$.Given $\lambda_1 = 1.32 \ \mathring{A}$,$Z_1 = 78$,$\lambda_2 = 4.17 \ \mathring{A}$,and $b = 7.4$.Substituting the values: $\sqrt{\frac{4.17}{1.32}} = \frac{78 - 7.4}{Z_2 - 7.4}$.$\sqrt{3.159} \approx 1.777 = \frac{70.6}{Z_2 - 7.4}$.$Z_2 - 7.4 = \frac{70.6}{1.777} \approx 39.73$.$Z_2 \approx 39.73 + 7.4 = 47.13$.Rounding to the nearest integer,the atomic number $Z_2$ is $47$.

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