If the wavelenth of K _ α radiation emitted by an atom of atomic...

$\dfrac{1}{λ} ∝ \left(\text{Z}\text{-1}\right)2{}^{}$

$∴ \dfrac{\lambda1}{\lambda2} = \dfrac{\left({Z}_{2} - 1\right)2}{\left({Z}_{1} - 1\right)2}$

$\dfrac{1}{4} = \left(\dfrac{{Z}_{2} - 1}{41 - 1}\right)2$

$\dfrac{1}{2} = \dfrac{{Z}_{2} - 1}{40}$

Solving this, we get, ${Z}_{2} = 21$