The wavelength of Kα, $$X-rays$$ produced by an $$X-ray$$ tube is $$0.76$$ Ao. The atomic number of the anode material of the tube is $$________$$.

We use $$1$$λ$$=R(Z$$−$$1)21$$−$$14$$⇒$$Z=43$$λ$$R+1=43$$×$$0.76$$×$$10$$−$$10$$×$$10967800+1=41$$

X-rays

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Question

The wavelength of $K_{α}$ , X-rays produced by an X-ray tube is 0.76 $\overset{o}{A}$ . The atomic number of the anode material of the tube is ________.

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Solution

$\begin{array}{l} We use \frac{1}{λ} = R (Z - 1)^{2} [1 - \frac{1}{4}] \\ \Rightarrow Z = \sqrt{\frac{4}{3 λ R}} + 1 = \sqrt{\frac{4}{3 \times 0.76 \times 10^{- 10} \times 10967800}} + 1 = 41 \end{array}$

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