If the wavelength of the K_alpha line of Moly | vedclass

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(B) According to Moseley's law,the frequency of the $K_\alpha$ line is given by $
u = cR(Z-1)^2(\frac{1}{1^2} - \frac{1}{2^2}) = \frac{3}{4}cR(Z-1)^2

Vedclass · Accepted Answer

$1.414\, \mathring A$

Vedclass · Answer

(B) According to Moseley's law,the frequency of the $K_\alpha$ line is given by $
u = cR(Z-1)^2(\frac{1}{1^2} - \frac{1}{2^2}) = \frac{3}{4}cR(Z-1)^2$.Since $\lambda = \frac{c}{
u}$,we have $\frac{1}{\lambda} \propto (Z-1)^2$,which implies $\lambda \propto \frac{1}{(Z-1)^2}$.Therefore,$\frac{\lambda_{Zn}}{\lambda_{Mo}} = \frac{(Z_{Mo}-1)^2}{(Z_{Zn}-1)^2} = \left(\frac{42-1}{30-1}ight)^2 = \left(\frac{41}{29}ight)^2$.$\lambda_{Zn} = \lambda_{Mo} 	imes \left(\frac{41}{29}ight)^2 = 0.7078 	imes \left(\frac{1681}{841}ight) \approx 0.7078 	imes 1.9988 \approx 1.414\, \mathring A$.

If the wavelength of the $K_\alpha$ line of Molybdenum (atomic number $Z = 42$) is $0.7078\, \mathring A$,then the wavelength of the $K_\alpha$ line of Zinc (atomic number $Z = 30$) is .......

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