How does the wavelength lambda of the K_alpha | vedclass

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(D) According to Moseley's Law for the characteristic $X$-ray spectrum,the frequency $
u$ of the $K_\alpha$ line is given by $
u = a(Z - b)^2$,where

Vedclass · Accepted Answer

$\lambda \propto \frac{1}{(Z - 1)^2}$

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(D) According to Moseley's Law for the characteristic $X$-ray spectrum,the frequency $
u$ of the $K_\alpha$ line is given by $
u = a(Z - b)^2$,where $a$ and $b$ are constants.For the $K_\alpha$ line,the screening constant $b = 1$.Since the frequency $
u = \frac{c}{\lambda}$,where $c$ is the speed of light,we have $\frac{c}{\lambda} = a(Z - 1)^2$.Rearranging this for the wavelength,we get $\lambda = \frac{c}{a(Z - 1)^2}$.Therefore,$\lambda \propto \frac{1}{(Z - 1)^2}$.

How does the wavelength $\lambda$ of the $K_\alpha$ line in the $X$-ray spectrum depend on the atomic number $Z$?

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