Question

X - rays emitted from a copper target and a molybdenum target are found to contain a line of wavelength 22.85 nm attributed to the Ka line of an impurity element. The Ka lines of copper (Z = 29) and molybdenum (Z = 42) have wavelength 15.42 nm and 15.42 nm and 7.12 nm respectively. Using Moseley's law ${\gamma }^{\frac{1}{2}}=a(Z-b)$. Calculate the atomic number of the impurity element.

Answer 1

$\gamma =28.85\times {10}^{-9}m$
copper (29) $=15.42\times {10}^{-9}m$
Molybdenum $\gamma =7.12\times {10}^{-9}m$
$y^{\frac{1}{2}}=9(Z-b)$
$(22.85\times {10}^9{)}^{\frac{1}{2}}=(29(Z-42){)}^2\Rightarrow 22.85\times {10}^{-9}=841Z^2-51156\Rightarrow 22.85\times {10}^{-9}+51.156\times {10}^{-3}=84$