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Question

(A) According to Moseley's Law,the frequency of characteristic $X$-rays is given by $
u = a(Z - b)^2$,where $b = 1$ for $K_\alpha$ lines.Thus,$
u \p

Vedclass · Accepted Answer

$16/25$

Vedclass · Answer

(A) According to Moseley's Law,the frequency of characteristic $X$-rays is given by $
u = a(Z - b)^2$,where $b = 1$ for $K_\alpha$ lines.Thus,$
u \propto (Z - 1)^2$.For $Z_1 = 64$,$
u_\alpha \propto (64 - 1)^2 = 63^2$.For $Z_2 = 80$,$
u'_\alpha \propto (80 - 1)^2 = 79^2$.The ratio is $\frac{
u_\alpha}{
u'_\alpha} = \frac{(63)^2}{(79)^2} = \frac{3969}{6241}$.Note: If the question assumes the simplified relation $
u \propto Z^2$ (ignoring the screening constant $b=1$),then $\frac{
u_\alpha}{
u'_\alpha} = \frac{64^2}{80^2} = (\frac{64}{80})^2 = (\frac{4}{5})^2 = \frac{16}{25}$.

If the frequency of the $K_\alpha$ line of an element with $Z = 64$ is $\nu_\alpha$ and the frequency of the $K_\alpha$ line of an element with $Z = 80$ is $\nu'_\alpha$,then the ratio of their frequencies is:

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