Question

If ${\lambda }_{\text{cu }}$ is the wavelength of $K_{\alpha }X$-ray line of copper (atomic number 29 ) and ${\lambda }_{Mo}$ 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 $n_1=2to{n}_2=1$
$\therefore \frac{1}{\lambda }=R(Z-b{)}^2[\frac{1}{(1{)}^2}-\frac{1}{(2{)}^2}]$
For K-series, b - 1
$\therefore \frac{1}{\lambda }\propto (Z-1{)}^2$
$\Rightarrow \frac{{\lambda }_{Cu}}{{\lambda }_{Mo}}=\frac{(Z_{Mo}-1{)}^2}{(Z_{Cu}-1{)}^2}=\frac{(42-1{)}^2}{(29-1{)}^2}$
$=\frac{41\times 41}{28\times 28}=\frac{1681}{784}=2.144$