If the frequency of K_alpha X-rays emitted fr | vedclass

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(C) According to Moseley's Law,the frequency $v$ of characteristic $X$-rays is given by $\sqrt{v} = a(Z - b)$,where $Z$ is the atomic number and $b$ i

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$\frac{25}{9} v$

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(C) According to Moseley's Law,the frequency $v$ of characteristic $X$-rays is given by $\sqrt{v} = a(Z - b)$,where $Z$ is the atomic number and $b$ is the screening constant. For $K_\alpha$ $X$-rays,$b = 1$.For the first element with $Z_1 = 31$,the frequency is $v_1 = v$:$\sqrt{v} = a(31 - 1) = 30a \quad \dots (i)$For the second element with $Z_2 = 51$,let the frequency be $v_2$:$\sqrt{v_2} = a(51 - 1) = 50a \quad \dots (ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{\sqrt{v_2}}{\sqrt{v}} = \frac{50a}{30a} = \frac{5}{3}$Squaring both sides:$\frac{v_2}{v} = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$Therefore,$v_2 = \frac{25}{9} v$.

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:

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