In X-ray spectrum,the wavelength lambda of th | vedclass

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(D) According to Moseley's Law,the frequency $f$ of the characteristic $X$-rays is given by $f = a(Z - b)^2$.For the $K_{\alpha}$ line,the screening c

Vedclass · Accepted Answer

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

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(D) According to Moseley's Law,the frequency $f$ of the characteristic $X$-rays is given by $f = a(Z - b)^2$.For the $K_{\alpha}$ line,the screening constant $b = 1$,so $f \propto (Z - 1)^2$.Since the frequency $f$ is inversely proportional to the wavelength $\lambda$ $(f = \frac{c}{\lambda})$,we have $\frac{c}{\lambda} \propto (Z - 1)^2$.Therefore,$\lambda \propto \frac{1}{(Z - 1)^2}$.

In $X$-ray spectrum,the wavelength $\lambda$ of the $K_{\alpha}$ line depends on the atomic number $Z$ as:

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