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(D) According to Moseley's Law,the frequency of characteristic $X$-rays is given by $
u = a(Z - b)^2$. For $K_{\alpha}$ lines,$b = 1$,so $
u \propto

Vedclass · Accepted Answer

$\frac{25}{16}$

Vedclass · Answer

(D) According to Moseley's Law,the frequency of characteristic $X$-rays is given by $
u = a(Z - b)^2$. For $K_{\alpha}$ lines,$b = 1$,so $
u \propto (Z - 1)^2$.Since $\lambda = \frac{c}{
u}$,we have $\lambda \propto \frac{1}{(Z - 1)^2}$.Given $Z_1 = 64$ and $Z_2 = 80$,the ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{(Z_2 - 1)^2}{(Z_1 - 1)^2}$.Substituting the values: $\frac{\lambda_1}{\lambda_2} = \frac{(80 - 1)^2}{(64 - 1)^2} = \frac{79^2}{63^2} \approx \left(\frac{80}{64}ight)^2 = \left(\frac{5}{4}ight)^2 = \frac{25}{16}$.Thus,the ratio is $\frac{25}{16}$.

When the atomic number of the first target is $Z_1 = 64$ and the atomic number of the second target is $Z_2 = 80$,the approximate ratio of the wavelengths of the $K_{\alpha,1}$ and $K_{\alpha,2}$ radiations is .......

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