The wavelength of the K_alpha line for an ele | vedclass

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

Vedclass · Accepted Answer

$4\lambda$

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(C) According to Moseley's law,the frequency $
u$ of the $K_\alpha$ line is given by $
u = cR(Z-b)^2(\frac{1}{1^2} - \frac{1}{2^2})$,which implies $\frac{1}{\lambda} \propto (Z-b)^2$.For the first element with $Z_1 = 29$ and $b = 1$,we have $\frac{1}{\lambda_1} = k(29-1)^2 = k(28)^2$,where $\lambda_1 = \lambda$.For the second element with $Z_2 = 15$ and $b = 1$,we have $\frac{1}{\lambda_2} = k(15-1)^2 = k(14)^2$.Taking the ratio of the two equations:$\frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - b)^2}{(Z_2 - b)^2} = \frac{(29-1)^2}{(15-1)^2} = \frac{28^2}{14^2} = (\frac{28}{14})^2 = 2^2 = 4$.Therefore,$\lambda_2 = 4\lambda_1 = 4\lambda$.

The wavelength of the $K_\alpha$ line for an element of atomic number $Z_1 = 29$ is $\lambda$. What is the wavelength of the $K_\alpha$ line for an element of atomic number $Z_2 = 15$? (Take Moseley's constant $b = 1$ for both elements.)

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