A cobalt target is bombarded with electrons and the wavelength of its characteristic spectrum is mea — Atomic Structure Chemistry Question
Question
A cobalt target is bombarded with electrons and the wavelength of its characteristic spectrum is measured. A second fainter characteristic spectrum is also found, due to an impurity in the target. The wavelengths of the $K_\alpha$ lines are 180.0 pm (cobalt, Z = 27) and 144.0 pm (impurity). The atomic number of impurity is
💡 Solution & Explanation
According to Moseley's law, $\frac{1}{\lambda} \propto (Z-1)^2$. Thus, $\frac{\lambda_{imp}}{\lambda_{Co}} = \frac{(Z_{Co}-1)^2}{(Z_{imp}-1)^2}$. Substituting values: $\frac{144.0}{180.0} = \frac{(27-1)^2}{(Z_{imp}-1)^2} \implies \frac{4}{5} = \frac{26^2}{(Z_{imp}-1)^2}$. Solving for $(Z_{imp}-1)^2$ gives $676 \times \frac{5}{4} = 845$. Taking the square root, $Z_{imp}-1 \approx 29.06 \implies Z_{imp} \approx 30$. Therefore, correct answer is 0030.