Question

Moseley's law for $K_{\alpha}$ photon is given by $\sqrt{v}=a(Z-b)$ where $a$ is a universal constant and $b$ is a screening constant. Moseley's logic helps us assume $b=1$ for $K_{\alpha}$ photon. If there is a percentage error $=10^{-2}$ in the measurement of $b$ due to actual orbital configuration of an atom, then the relative error in the measurement of $v$ if $Z=51$ is $p \times 10^{-q}$. Find the value of $(p q)$.

Answer 1

$\frac{d b}{b}=10^{-4}$
$ \Rightarrow d b=10^{-4}$
$\frac{1}{2} \frac{d v}{v}=\frac{d(Z-b)}{(Z-b)} $
$\Rightarrow \frac{d v}{v}=\frac{2 \times 10^{-4}}{50}$
$=4 \times 10^{-6}$