If lambda_{Cu} is the wavelength of the K_{al | vedclass

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Question

(B) According to Moseley's law,the frequency $
u$ of the $K_{\alpha}$ $X$-ray line is given by $\sqrt{
u} = a(Z - b)$,where $b = 1$ for the $K_{\alp

Vedclass · Accepted Answer

$2.14$

Vedclass · Answer

(B) According to Moseley's law,the frequency $
u$ of the $K_{\alpha}$ $X$-ray line is given by $\sqrt{
u} = a(Z - b)$,where $b = 1$ for the $K_{\alpha}$ transition.Since $
u = \frac{c}{\lambda}$,we have $\sqrt{\frac{c}{\lambda}} = a(Z - 1)$,which implies $\frac{1}{\sqrt{\lambda}} \propto (Z - 1)$.Therefore,the ratio of the wavelengths is given by $\frac{\sqrt{\lambda_{Mo}}}{\sqrt{\lambda_{Cu}}} = \frac{Z_{Cu} - 1}{Z_{Mo} - 1}$.Squaring both sides,we get $\frac{\lambda_{Mo}}{\lambda_{Cu}} = \left( \frac{29 - 1}{42 - 1} ight)^2 = \left( \frac{28}{41} ight)^2$.Thus,the ratio $\frac{\lambda_{Cu}}{\lambda_{Mo}} = \left( \frac{41}{28} ight)^2 = \frac{1681}{784} \approx 2.144$.Therefore,the ratio is close to $2.14$.

If $\lambda_{Cu}$ is the wavelength of the $K_{\alpha}$ $X$-ray line of copper (atomic number $Z = 29$) and $\lambda_{Mo}$ is the wavelength of the $K_{\alpha}$ $X$-ray line of molybdenum (atomic number $Z = 42$),then the ratio $\frac{\lambda_{Cu}}{\lambda_{Mo}}$ is close to:

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