Question

The wavelength $K_{\alpha }X$-rays produced by an $X$-ray tube is $0.76\mathring{A}$. The atomic number of anticathode material is

• A. 82
• B. 41
• C. 20
• D. 10

Answer 1

Answer: (B)

For $K_{\alpha }X$-ray line
$\frac{1}{{\lambda }_{\alpha }}=R(Z-1{)}^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3R}{4}(Z-1{)}^2$
On putting the given values
$\frac{1}{0.76\times 10^{-10}}=\frac{3}{4}\times 1.09\times 10^7(Z-1{)}^2$
$\Rightarrow (Z-1{)}^2\approx 1600$
$\Rightarrow Z-1=40$
$\Rightarrow Z=41$