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

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(A) The wavelength of the $L_{\alpha}$ line is given by Moseley's law for $X$-rays:$\frac{1}{\lambda} = R(Z - b)^2 \left( \frac{1}{n_1^2} - \frac{1}{n

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$5.41$

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(A) The wavelength of the $L_{\alpha}$ line is given by Moseley's law for $X$-rays:$\frac{1}{\lambda} = R(Z - b)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$For the $L_{\alpha}$ line,$n_1 = 2$ and $n_2 = 3$,and the screening constant $b \approx 7.4$.Thus,$\frac{1}{\lambda} \propto (Z - 7.4)^2$,which implies $\lambda \propto \frac{1}{(Z - 7.4)^2}$.Given $\lambda_1 = 1.30 \ \mathring{A}$ for $Z_1 = 78$ and $\lambda_2$ for $Z_2 = 42$:$\frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - 7.4)^2}{(Z_2 - 7.4)^2}$$\lambda_2 = 1.30 	imes \left( \frac{78 - 7.4}{42 - 7.4} ight)^2$$\lambda_2 = 1.30 	imes \left( \frac{70.6}{34.6} ight)^2$$\lambda_2 = 1.30 	imes (2.04)^2 \approx 1.30 	imes 4.16 = 5.41 \ \mathring{A}$.

The $X$-ray wavelength of the $L_{\alpha}$ line of platinum $(Z=78)$ is $1.30 \ \mathring{A}$. The $X$-ray wavelength of the $L_{\alpha}$ line of Molybdenum $(Z=42)$ is ............. $\mathring{A}$.

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