Question

The frequency of $K_{\alpha }$ $X-$ray emitted from the element with atomic number $31$ is $f$. If the frequency of $K_{\alpha }$ $X-$ray emitted from the element with atomic number $51$ is given by $\frac{25}{x}f$, then the value of $x$ is ________.

$($Assume that the value of screening constant for $K_{\alpha }$ is $1)$

Answer 1

According to Moseley's law, $\sqrt{\nu }=a(Z-1)$

For the first element with atomic number $31$,

$\sqrt{f}=a(31-1)......\left(1\right)$

Similarly, for the second element with atomic number $51$,

$\sqrt{\frac{25f}{x}}=a(51-1)......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get,

$x=9$