Question 40.40: (a) From Eq. 40-26, what is the ratio of the photon energies......
(a) From Eq. 40-26, what is the ratio of the photon energies due to K_\alpha transitions in two atoms whose atomic numbers are Z and Z′? (b) What is this ratio for uranium and aluminum? (c) For uranium and lithium?
\begin{aligned} f & =\frac{\Delta E}{h}=\frac{(10.2 eV )(Z-1)^2}{\left(4.14 \times 10^{-15} eV \cdot s \right)} \\ & =\left(2.46 \times 10^{15} Hz \right)(Z-1)^2 . \end{aligned} (40-26)
(a) According to Eq. 40-26, f \propto(Z-1)^2 , so the ratio of energies is (using Eq. 38-2)
E=h f \quad \text { (photon energy) } (38-2)
\frac{f}{f^{\prime}}=\left(\frac{Z-1}{Z^{\prime}-1}\right)^2
(b) We refer to Appendix F. Applying the formula from part (a) to Z = 92 and Z’ = 13, we obtain
\frac{E}{E^{\prime}}=\frac{f}{f^{\prime}}=\left(\frac{Z-1}{Z^{\prime}-1}\right)^2=\left(\frac{92-1}{13-1}\right)^2=57.5 .
(c) Applying this to Z = 92 and Z’ = 3, we obtain
\frac{E}{E^{\prime}}=\left(\frac{92-1}{3-1}\right)^2=2.07 \times 10^3 .