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

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(D) According to Moseley's Law for $K_{\alpha}$ $X$-ray emission,the frequency $
u$ is given by $
u \propto (Z - 1)^2$.Since $\lambda = c/
u$,we ha

Vedclass · Accepted Answer

$4\lambda$

Vedclass · Answer

(D) According to Moseley's Law for $K_{\alpha}$ $X$-ray emission,the frequency $
u$ is given by $
u \propto (Z - 1)^2$.Since $\lambda = c/
u$,we have $\frac{1}{\lambda} \propto (Z - 1)^2$,which implies $\lambda \propto \frac{1}{(Z - 1)^2}$.Given $\lambda_1 = \lambda$ for $Z_1 = 57$ and we need to find $\lambda_2$ for $Z_2 = 29$.Using the ratio: $\frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - 1)^2}{(Z_2 - 1)^2}$.Substituting the values: $\frac{\lambda_2}{\lambda} = \frac{(57 - 1)^2}{(29 - 1)^2} = \frac{56^2}{28^2}$.$\frac{\lambda_2}{\lambda} = (\frac{56}{28})^2 = 2^2 = 4$.Therefore,$\lambda_2 = 4\lambda$.

The wavelength of the $K_{\alpha}$ line for an element with atomic number $57$ is $\lambda$. What will be the wavelength of the $K_{\alpha}$ line for an element with atomic number $29$?

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