Moseley's equation is represented as $\sqrt{\mathrm{v}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})$ where, a and bare constants. 

If OA = 1, then atomic number of the element showing frequency of 400 Hz is

$\sqrt{\mathrm{v}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})=\mathrm{aZ}-\mathrm{ab}$

Thus, graph is a straight line

$\sqrt{\mathrm{v}}=0 \mathrm{aZ}=\mathrm{ab}$

                             $\mathrm{Z}=\mathrm{b}=\mathrm{OA}=1$

$\text{ Thus, }\mathrm{b}=1\text{ and at }\mathrm{Z}=0,\sqrt{\mathrm{v}}=-\mathrm{ab}$

            $\text{ Slope }=\mathrm{a}=\tan {45}^{\circ }=1$

$\mathrm{Thus} \sqrt{\mathrm{v}}=(\mathrm{Z}-1)$

If v = 400 Hz, then

              $\begin{array}{l} \sqrt{400}=\mathrm{Z}-1\\ 20=\mathrm{Z}-1\\ \therefore \mathrm{Z}=21\end{array}$