If the frequency of ${K_\alpha }$ X rays emitted from the element with atomic number 31 is v, then the frequency of ${K_\alpha }$ X-rays emitted from the element with atomic number 51 would be
(A) $\dfrac{5}{3}v$
(B) $\dfrac{{54}}{{31}}v$
(C) $\dfrac{{25}}{9}v$
(D) $\dfrac{9}{{25}}v$

Hint We write down the known formula of Moseley’s law $\sqrt \nu \propto \left( {z - 1} \right)$ . Then we will square both the sides and will take the ratio. The constant proportionality term will get cancelled. We will easily get the unknown value of frequency.

Complete step by step solution
We know that the statement of Moseley’s law that the square root of the frequency of a peak of the characteristics X- ray spectrum of any element is directly proportional to the atomic number of that element.
Here let the frequency of the ${K_\alpha }$ line of any element having atomic number Z be ${\nu _1}$ . Then according to Moseley’s law.
Given that ${\nu _1} = \nu $
When frequency is v then atomic number be ${Z_1} = 31$
Now when atomic number ${Z_2} = \,51$ then let frequency of ${K_\alpha }$ be ${\nu _2}$
 $\sqrt \nu \propto \left( {z - 1} \right)$
So squaring both sides and taking ratio we can write
 $ \Rightarrow \dfrac{{{\nu _2}}}{{{\nu _1}}} = {\left( {\dfrac{{{Z_2} - 1}}{{{Z_1} - 1}}} \right)^2}$
So, we have to calculate ${\nu _2}$
 $ \Rightarrow {\nu _2} = {\left( {\dfrac{{51 - 1}}{{31 - 1}}} \right)^2}{\nu _1} = \,\,\dfrac{{25}}{9}\nu $

Thus, the required solution is $\dfrac{{25}}{9}\nu $ (option- “C”)

Additional Information
Moseley's law can be explained from Bohr’s theory. From Bohr’s theory we can write that the equation for n-th energy state of the atom is ${E_n} = - \dfrac{{13.6{{(z - 1)}^2}}}{{{n^2}}}$ (in eV). Due to the ${K_\alpha }$ of spectrum is produced due to the transition of an electron from \[L\left( {n = 2} \right)\] orbit to \[K\left( {n = 1} \right)\] orbit. Due to transition if the frequency of the emitted X-rays be v
Then $h\nu = \,{(E)_{n = 2}} - {(E)_{n = 1}} = {E_2} - {E_1}$

 $ \Rightarrow h\nu = \, - \dfrac{{13.6{{(z - 1)}^2}}}{{{2^2}}} + \dfrac{{13.6{{(z - 1)}^2}}}{{{1^2}}}\,(in\,eV)$
 $\therefore h\nu = 10.2{(z - 1)^2}$
 $\sqrt \nu \propto \left( {z - 1} \right)$

Note
Solving this one may have to keep in mind that the square root of the frequency is of a peak of the characteristics X- ray spectrum of any element is directly proportional to the atomic number of that element. So, option “D” cannot be correct. If one quantity increases other ones also increases. Here we are not going to use proportionality constant to calculate the unknown value of frequency. When we take a ratio it automatically gets cancelled. We have to square in the last step so option “D” also cannot be the right one. Option “C” is the correct one.