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  4. The Kα line obtained for molybdenum (Z=42) is 0.71 mathringA. Then, the wavelength of the Kα line of copper (Z=29) is

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Q. The $K_{\alpha}$ line obtained for molybdenum $(Z=42)$ is $0.71\,\mathring{A}$. Then, the wavelength of the $K_{\alpha}$ line of copper $(Z=29)$ is

NTA AbhyasNTA Abhyas 2022

Solution:

$\lambda_{\min } \propto \frac{1}{Z-1^{2}}$
$\therefore \frac{\lambda_{ Cu }}{\lambda_{ Mo }}=\frac{z_{ Mo }-1^{2}}{z_{ Cu }-1^{2}}=\frac{41^{2}}{28^{2}}=2.14$
$\therefore \lambda_{ Cu }=2.14 \times 0.71 \,\mathring{A}=1.52\,\mathring{A}$