Question

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

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

Answer 1

The correct option is A $Z=6$

From Moseley's law,

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

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

$\Rightarrow \frac{4}{1}=\frac{{10}^2}{(Z_2-1{)}^2}\Rightarrow Z_2=6$

$\therefore Z_2=6$

Hence, option $\left(\text{A}\right)$ is correct.