If the K_alpha radiation of Mo (Z=42) has a w | vedclass

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(B) According to Moseley's Law,the frequency of characteristic $X$-rays is given by $
u = a(Z-b)^2$. Since $
u = c/\lambda$,we have $\lambda \propto

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(B) According to Moseley's Law,the frequency of characteristic $X$-rays is given by $
u = a(Z-b)^2$. Since $
u = c/\lambda$,we have $\lambda \propto 1/(Z-1)^2$ for $K_\alpha$ radiation.Given for $Mo (Z_1 = 42)$,$\lambda_1 = 0.71 \;\mathring A$.For $Cu (Z_2 = 29)$,we need to find $\lambda_2$.Using the ratio: $\frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - 1)^2}{(Z_2 - 1)^2}$.Substituting the values: $\lambda_2 = 0.71 	imes \frac{(42 - 1)^2}{(29 - 1)^2}$.$\lambda_2 = 0.71 	imes \frac{41^2}{28^2} = 0.71 	imes \frac{1681}{784}$.$\lambda_2 \approx 0.71 	imes 2.144 = 1.522 \;\mathring A$.Thus,the wavelength is $1.52 \;\mathring A$.

If the $K_\alpha$ radiation of $Mo (Z=42)$ has a wavelength of $0.71 \;\mathring A$,calculate the wavelength of the corresponding $K_\alpha$ radiation of $Cu (Z=29)$ (in $\mathring A$). (in $.52$)

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