The wavelength lambda of the {K_{alpha}} line | vedclass

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(C) According to Moseley's law,the frequency $
u$ of characteristic $X$-rays is given by $
u = a(Z - b)^2$,where $a$ and $b$ are constants.Since the

Vedclass · Accepted Answer

$\lambda \propto \frac{1}{Z^2}$

Vedclass · Answer

(C) According to Moseley's law,the frequency $
u$ of characteristic $X$-rays is given by $
u = a(Z - b)^2$,where $a$ and $b$ are constants.Since the frequency $
u$ is related to the wavelength $\lambda$ by the relation $
u = \frac{c}{\lambda}$,we have $\frac{c}{\lambda} \propto (Z - b)^2$.For large values of $Z$,the constant $b$ can be neglected,leading to $\frac{1}{\lambda} \propto Z^2$.Therefore,the wavelength $\lambda$ varies with the atomic number $Z$ as $\lambda \propto \frac{1}{Z^2}$.

The wavelength $\lambda$ of the ${K_{\alpha}}$ line of characteristic $X$-ray spectra varies with atomic number $Z$ approximately as:

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