The wavelength of the K_{alpha} X-ray produce | vedclass

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(C) According to Moseley's Law for $K_{\alpha}$ $X$-rays:$\frac{1}{\lambda} = R(Z - 1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$For $K_{\alp

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$41$

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(C) According to Moseley's Law for $K_{\alpha}$ $X$-rays:$\frac{1}{\lambda} = R(Z - 1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$For $K_{\alpha}$ transition,$n_1 = 1$ and $n_2 = 2$.So,$\frac{1}{\lambda} = R(Z - 1)^2 \left( 1 - \frac{1}{4} ight) = R(Z - 1)^2 \left( \frac{3}{4} ight)$.Given $\lambda = 0.76 \ \mathring{A} = 0.76 	imes 10^{-10} \ m$ and Rydberg constant $R \approx 1.097 	imes 10^7 \ m^{-1}$.$(Z - 1)^2 = \frac{4}{3 \lambda R} = \frac{4}{3 	imes 0.76 	imes 10^{-10} 	imes 1.097 	imes 10^7}$.$(Z - 1)^2 \approx \frac{4}{2.501} 	imes 10^3 \approx 1600$.$Z - 1 = \sqrt{1600} = 40$.$Z = 41$.

The wavelength of the $K_{\alpha}$ $X$-ray produced by an $X$-ray tube is $0.76 \ \mathring{A}$. The atomic number of the anode material is ...

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