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  4. The wavelength Kα X-rays produced by an X-ray tube is 0.76 mathringA. The atomic number of anticathode material is

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

Atoms

Solution:

For $K_{\alpha} X$-ray line
$\frac{1}{\lambda_{\alpha}}=R(Z-1)^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=\frac{3 R}{4}(Z-1)^{2}$
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$