Question

$\text{X}-$ ray from a tube with a target $\text{A}$ of atomic number $Z$ shows strong $K-$ lines for target $\text{A}$ and weak $K-$ lines for impurities. The wavelength of $K_{\alpha }$ lines is ${\lambda }_{\alpha }$ for target $\text{A}$ and ${\lambda }_1$ and ${\lambda }_2$ for two impurities. $$\frac{{\lambda }_{\alpha }}{{\lambda }_1}=4\text{and}\frac{{\lambda }_{\alpha }}{{\lambda }_2}=\frac{1}{4}$$Assuming the screening constant of $K_{\alpha }$ lines to be unity. The atomic number of the impurities $Z_1$ and $Z_2$ is

• A. $Z_1=2Z-1;Z_2=\frac{Z+1}{2}$
• B. $Z_1=2Z+1;Z_2=\frac{Z}{2}+1$
• C. $Z_1=\frac{Z+1}{2};Z_2=2Z-1$
• D. $Z_1=\frac{Z}{2}+1;Z_2=2Z+1$

Answer 1

The correct option is A $Z_1=2Z-1;Z_2=\frac{Z+1}{2}$

Given, $\frac{{\lambda }_{\alpha }}{{\lambda }_1}=4;\frac{{\lambda }_{\alpha }}{{\lambda }_2}=\frac{1}{4}$

According to moseley's law,

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

$\Rightarrow \frac{{\lambda }_{\alpha }}{{\lambda }_1}=\frac{(Z_1-1{)}^2}{(Z-1{)}^2}$

From the data given in the question,

$\frac{Z_1-1}{Z-1}=2\Rightarrow Z_1=2Z-1$

Similarly, $\frac{{\lambda }_{\alpha }}{{\lambda }_2}=\frac{(Z_2-1{)}^2}{(Z-1{)}^2}$

From the data given in the question,

$\frac{Z_2-1}{Z-1}=\frac{1}{2}\Rightarrow Z_2=\frac{Z+1}{2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.