Question

The $K_{\alpha }$ line obtained for molybdenum $(Z=42)$ is $0.71\mathring{A}$. Then, the wavelength of the $K_{\alpha }$ line of copper $(Z=29)$ is

• A. $2.14\mathring{A}$
• B. $1.52\mathring{A}$
• C. $1.04\mathring{A}$
• D. $0.71\mathring{A}$

Answer 1

Answer: (B)

${\lambda }_{\min }\propto \frac{1}{Z-1^2}$
$\therefore \frac{{\lambda }_{Cu}}{{\lambda }_{Mo}}=\frac{z_{Mo}-1^2}{z_{Cu}-1^2}=\frac{41^2}{28^2}=2.14$
$\therefore {\lambda }_{Cu}=2.14\times 0.71\mathring{A}=1.52\mathring{A}$