The wavelength of the L_alpha X-ray line of P | vedclass

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(B) The wavelength of the $L_\alpha$ line is given by Moseley's law: $\frac{1}{\lambda} = R(Z-b)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$.Takin

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(B) The wavelength of the $L_\alpha$ line is given by Moseley's law: $\frac{1}{\lambda} = R(Z-b)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$.Taking the ratio for two elements:$\frac{\lambda_2}{\lambda_1} = \left( \frac{Z_1 - b}{Z_2 - b} \right)^2$.Given $\lambda_1 = 1.32 \, \mathring A$,$Z_1 = 78$,$\lambda_2 = 4.17 \, \mathring A$,$b = 7.4$.$\frac{4.17}{1.32} = \left( \frac{78 - 7.4}{Z_2 - 7.4} \right)^2$.$3.159 \approx \left( \frac{70.6}{Z_2 - 7.4} \right)^2$.Taking the square root: $1.777 \approx \frac{70.6}{Z_2 - 7.4}$.$Z_2 - 7.4 \approx \frac{70.6}{1.777} \approx 39.73$.$Z_2 \approx 39.73 + 7.4 = 47.13$.Rounding to the nearest integer,the atomic number is $47$.

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