Modern Physics 4 Question 27
28. The wavelength of $K _{\alpha}, X$-rays produced by an X-ray tube is $0.76 \AA$. The atomic number of the anode material of the tube is
(1996, 2M)
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Solution:
- $\frac{1}{\lambda _{K _{\alpha}}}=R(Z-1)^{2} \frac{1}{1^{2}}-\frac{1}{2^{2}}=\frac{3}{4} R(Z-1)^{2}$
$$ \therefore \quad(Z-1)=\frac{2}{\sqrt{3 R \lambda _{K _{\alpha}}}} $$
Here, $R=$ Rydberg’s constant
$$ =1.097 \times 10^{7} m^{-1} \text { and } $$
$$ \lambda _{K _{\alpha}}=0.76 \AA=0.76 \times 10^{-10} m $$
Substituting the values, we have
$$ \text { or } \quad Z=39 $$
