Question

$K_{\alpha }$ wavelength emitted by an atom of atomic number Z = 11 is $\lambda$. The atomic number for an atom that emits $K_{\alpha }$ radiation with wavelength $4\lambda$ is

• A. Z = 6
• B. Z = 4
• C. Z = 11
• D. Z = 44

Answer 1

The correct option is A

Z = 6

$\frac{1}{\lambda }\alpha (Z-1{)}^2$

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\left(\frac{Z_2-1}{Z_1-1}or\frac{1}{4}\right)={\left(\frac{Z_2-1}{11-1}\right)}^2$

Solving this, we get, $Z_2=6$