Frequency of the K_{alpha} X-ray for the elem | vedclass

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Question

(A) According to Moseley's law,the frequency $
u$ of $K_{\alpha}$ $X$-rays is given by $
u = a(Z - b)^2$,where $a$ is a constant and $b = 1$ for $K_

Vedclass · Accepted Answer

$5.46 	imes 10^{18} \ Hz$

Vedclass · Answer

(A) According to Moseley's law,the frequency $
u$ of $K_{\alpha}$ $X$-rays is given by $
u = a(Z - b)^2$,where $a$ is a constant and $b = 1$ for $K_{\alpha}$ lines.Thus,$
u \propto (Z - 1)^2$.For calcium $(Ca, Z = 20)$,$
u_{Ca} = 8.95 	imes 10^{17} \ Hz$.For cadmium $(Cd, Z = 48)$,we have:$\frac{
u_{Cd}}{
u_{Ca}} = \left( \frac{Z_{Cd} - 1}{Z_{Ca} - 1} ight)^2$$\frac{
u_{Cd}}{8.95 	imes 10^{17}} = \left( \frac{48 - 1}{20 - 1} ight)^2 = \left( \frac{47}{19} ight)^2$$
u_{Cd} = 8.95 	imes 10^{17} 	imes \left( 2.4737 ight)^2$$
u_{Cd} = 8.95 	imes 10^{17} 	imes 6.119 = 5.476 	imes 10^{18} \ Hz$.Rounding to the nearest provided option,the frequency is $5.46 	imes 10^{18} \ Hz$.

Frequency of the $K_{\alpha}$ $X$-ray for the element calcium $(Ca, Z = 20)$ is $8.95 \times 10^{17} \ Hz$. The frequency of the $K_{\alpha}$ $X$-ray for the element cadmium $(Cd, Z = 48)$ will be:

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