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  4. Atomic number of anticathode material in an X-ray tube is 41 . Wavelength of Kα X-ray produced in the tube is

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Q. Atomic number of anticathode material in an $X$-ray tube is $41$ . Wavelength of $K_{\alpha} X$-ray produced in the tube is

Atoms

Solution:

$\sqrt{f}=K_{\alpha}(Z-1)$
$\sqrt{f}=(40)$
$\sqrt{f}=\sqrt{2.48 \times 10^{15}} \times 40$
$f=2.48 \times 10^{15} \times 1600$
$\frac{c}{\lambda}=2.48 \times 10^{15} \times 1600$
$\lambda=0.76 \times 10^{-10} $
or $\lambda=0.76 \,\mathring{A} $