Question

If $\lambda_{\text {cu }}$ is the wavelength of $K_{\alpha} X$-ray line of copper (atomic number 29 ) and $\lambda_{M o}$ is the wavelength of the $K_{\alpha}$ $X$-ray line of molybdenum (atomic number $42$ ), then the ratio $\lambda_{ cu } / \lambda_{ Mo }$ is close to

• A. 1.99
• B. 2.14
• C. 0.50
• D. 0.48

Answer 1

Answer: (B)

$K_{\alpha}$ transition takes place from $n_1 = 2\, \, to\, \, n_2 = 1$
$\therefore \, \, \, \frac{1}{\lambda} = R(Z -b)^2 \bigg [ \frac{1}{(1)^2} - \frac{1}{(2)^2} \bigg ]$
For K-series, b - 1
$\therefore \, \, \, \, \frac{1}{\lambda} \propto (Z - 1)^2$
$\Rightarrow \frac{\lambda_{Cu}}{\lambda_{Mo}} = \frac{(Z_{Mo} - 1)^2}{(Z_{Cu} - 1)^2} = \frac{(42 - 1)^2}{(29 - 1)^2}$
$\, \, \, \, \, = \frac{41 \times 41}{28 \times 28} = \frac{1681}{784} = 2.144$