Physics - Modern Physics Question with Solution | TestHub

Moseley's Law states that the frequency of characteristic X-rays is proportional to $(Z-b{)}^2$. For K$\alpha$ lines, $b=1$.

So, $\nu \propto (Z-1{)}^2$. Since $\lambda =c/\nu$, we have $\lambda \propto 1/(Z-1{)}^2$.

For iron (Fe), $Z_{Fe}=26$, ${\lambda }_{Fe}=193pm$.

For copper (Cu), $Z_{Cu}=29$.

$\frac{{\lambda }_{Cu}}{{\lambda }_{Fe}}=\frac{(Z_{Fe}-1{)}^2}{(Z_{Cu}-1{)}^2}$

${\lambda }_{Cu}={\lambda }_{Fe}{\left(\frac{Z_{Fe}-1}{Z_{Cu}-1}\right)}^2=193pm{\left(\frac{26-1}{29-1}\right)}^2=193pm{\left(\frac{25}{28}\right)}^2$

${\lambda }_{Cu}=193pm\times \frac{625}{784}\approx 193pm\times 0.797\approx 153.8pm$

Rounding to the nearest integer, ${\lambda }_{Cu}=154pm$.