The wavelength of the {K_{alpha}} line for an | vedclass

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(C) According to Moseley's law,the frequency of the ${K_{\alpha}}$ $X$-ray line is given by $
u = cR(Z-1)^2(\frac{1}{1^2} - \frac{1}{2^2}) = \frac{3}

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$\frac{9}{4}\lambda$

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(C) According to Moseley's law,the frequency of the ${K_{\alpha}}$ $X$-ray line is given by $
u = cR(Z-1)^2(\frac{1}{1^2} - \frac{1}{2^2}) = \frac{3}{4}cR(Z-1)^2$.Since $
u = \frac{c}{\lambda}$,we have $\frac{c}{\lambda} \propto (Z-1)^2$,which implies $\lambda \propto \frac{1}{(Z-1)^2}$.Given for $Z_1 = 43$,the wavelength is $\lambda_1 = \lambda$.For $Z_2 = 29$,let the wavelength be $\lambda_2$.Using the ratio: $\frac{\lambda_2}{\lambda_1} = \left(\frac{Z_1 - 1}{Z_2 - 1}ight)^2$.Substituting the values: $\frac{\lambda_2}{\lambda} = \left(\frac{43 - 1}{29 - 1}ight)^2 = \left(\frac{42}{28}ight)^2$.Simplifying the fraction: $\frac{42}{28} = \frac{3}{2}$.Therefore,$\frac{\lambda_2}{\lambda} = (\frac{3}{2})^2 = \frac{9}{4}$.Thus,$\lambda_2 = \frac{9}{4}\lambda$.

The wavelength of the ${K_{\alpha}}$ line for an element of atomic number $43$ is $\lambda$. Then the wavelength of the ${K_{\alpha}}$ line for an element of atomic number $29$ is:

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