The K_{alpha} wavelength emitted by an atom o | vedclass

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Question

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

Vedclass · Accepted Answer

$Z = 6$

Vedclass · Answer

(A) According to Moseley's law,the frequency of $K_{\alpha}$ $X$-rays is given by $
u = c/\lambda = a(Z - b)^2$,where $b = 1$ for $K_{\alpha}$ lines.Thus,$\frac{1}{\lambda} \propto (Z - 1)^2$,or $\frac{1}{\sqrt{\lambda}} \propto (Z - 1)$.For the first atom,$Z_1 = 11$ and $\lambda_1 = \lambda$. So,$\frac{1}{\sqrt{\lambda}} = k(11 - 1) = 10k$.For the second atom,$Z_2 = Z$ and $\lambda_2 = 4\lambda$. So,$\frac{1}{\sqrt{4\lambda}} = k(Z - 1) = \frac{1}{2\sqrt{\lambda}} = k(Z - 1)$.Dividing the two equations: $\frac{1/\sqrt{\lambda}}{1/(2\sqrt{\lambda})} = \frac{10k}{k(Z - 1)}$.$2 = \frac{10}{Z - 1}$.$Z - 1 = 5$,which gives $Z = 6$.

The $K_{\alpha}$ wavelength emitted by an atom of atomic number $Z = 11$ is $\lambda$. Find the atomic number for an atom that emits $K_{\alpha}$ radiation with wavelength $4\lambda$.

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