Question

The wavelength of $K_x$ radiation of $z=11$, atom $95$, then atomic no of atom was $K_{\alpha }$ radiation, wavelength is $4_{\lambda }$.

Answer 1

$K_{\alpha }radiationofz=11is\lambda \sqrt{F}=c\times (z-1)F=\frac{c}{\lambda }\lambda \alpha \frac{1}{F}F\alpha {(z-1)}^2\frac{1}{z}\alpha {(z-1)}^2\frac{1}{\lambda }=K{\left(10\right)}^2\frac{1}{4\lambda }=K{(z^1-1)}^{2}4{(z^1-1)}^2={\left(10\right)}^2{z}^1-1=5z^1=6$