Question

If frequency of $k_{\alpha }$ x - ray emitted from the element with atomic number $31$ is $f$, then frequency of $k_{\alpha }$ x - ray emitted from the element with atomic number $51$ would be (Assuming screening constant for $k_{\alpha }$ x - ray is $1$)

• A. $\frac{5}{3}f$
• B. $\frac{51}{31}f$
• C. $\frac{9}{25}f$
• D. $\frac{25}{9}f$

Answer 1

The correct option is D $\frac{25}{9}f$

According to Moseley's Law-
$\sqrt{f}=\sqrt{\frac{3Rc}{4}}(z-1)$
Where $f$ = frequency
$R$ = Rydberg constant
and $z$ = atomic number
for atomic no. - $31.$
So, $\sqrt{f}=\sqrt{\frac{3Rc}{4}}(31-1)....\left(1\right)$
for atomic no. - 51
$\sqrt{f^{\prime }}=\sqrt{\frac{3Rc}{4}}(51-1).....\left(2\right)$
from equation (1) and (2)
$\sqrt{\frac{f}{f^{\prime }}}=\frac{3}{5}$
$\frac{f}{f^{\prime }}=\frac{9}{25}$
$f^{\prime }=\frac{25}{9}f$