The frequency of the characteristic X ray of $K_{\alpha }$ line of metal target ‘M’ is 2500 $c{m}^{-1}$  and the graph between $\sqrt{v}$ vs ‘z’ is as follows. Then the atomic number of M is___

$v=A(Z-b{)}^2$

- Here, v is the frequency of the X-ray emission line and A and b are the constants which depend on the line type, like K, L, etc.

- Now, let us come to the solution. We have the graph and the frequency of the X Ray as $2500 {\mathrm{cm}}^{-1}$. Now, according to the Moseley's Law, we can write $\begin{array}{r}\sqrt{v}=k_1\left(Z-k_2\right)\\ \Rightarrow \sqrt{v}=k_{1}Z-k_1{k}_2\end{array}$

- On comparing this With the equation of a straight line that is, y=mx+c, we can relate the LHS and RHS as:

$\begin{array}{r}y=\sqrt{v}=\sqrt{2500}=50\\ x=z\\ m=k_1=\tan {45}^0=1\\ c=-k_1{k}_2\\ OA=1;-k_1{k}_2=1\\ \Rightarrow k_2=-1\end{array}$

- Now, let us use the above information in order to find the required atomic number.

$\begin{array}{r}\sqrt{\mathrm{v}}={\mathrm{k}}_1\mathrm{Z}-{\mathrm{k}}_1{\mathrm{k}}_2\\ \Rightarrow 50=1\times \mathrm{Z}+\left(1\right)(-1)\\ \Rightarrow 50=\mathrm{Z}-1\\ \Rightarrow \mathrm{Z}=51\end{array}$