Question

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

• A. $\text{2.14 }\overset{^\circ }{A}$
• B. $\text{1.52 }\overset{^\circ }{A}$
• C. $\text{1.04 }\overset{^\circ }{A}$
• D. $\text{0.71 }\overset{^\circ }{A}$

Answer 1

Answer: (B)

$\lambda_{\min } \propto \frac{1}{(Z-1)^{2}}$
$\therefore \frac{\lambda_{ Cu }}{\lambda_{ Mo }}=\frac{\left(Z_{ M _{0}}-1\right)^{2}}{\left(Z_{ Cu }-1\right)^{2}}=\frac{4 l ^{2}}{(28)^{2}}=2.14$
$\therefore \lambda_{ Cu }=2.14 \times 0.71 A =1.52 A$