Question 40.46: Determine the constant C in Eq. 40-27 to five significant fi......
Determine the constant C in Eq. 40-27 to five significant figures by finding C in terms of the fundamental constants in Eq.40-24 and then using data from Appendix B to evaluate those constants. Using this value of C in Eq. 40-27, determine the theoretical energy E_{theory} of the K_\alpha photon for the low-mass elements listed in the following table. The table includes the value (eV) of the measured energy E_{exp} of the K_\alpha photon for each listed element. The percentage deviation between E_{theory} and E_{exp} can be calculated as
\sqrt{f}=C Z-C, (40-27)
E_n=-\frac{m e^4}{8 \varepsilon_0^2 h^2} \frac{1}{n^2}=-\frac{13.60 eV }{n^2}, \quad \text { for } n=1,2,3, \ldots (40-24)
\text { percentage deviation }=\frac{E_{\text {theory }}-E_{\exp }}{E_{\exp }} 100 \text {. }
What is the percentage deviation for (a) Li, (b) Be, (c) B, (d) C, (e) N, (f) O, (g) F, (h) Ne, (i) Na, and (j) Mg?
\begin{array}{lllc} \hline Li & 54.3 & O & 524.9 \\ Be & 108.5 & F & 676.8 \\ B & 183.3 & Ne & 848.6 \\ C & 277 & Na & 1041 \\ N & 392.4 & Mg & 1254 \\ \hline \end{array}
(There is actually more than one K_\alpha ray because of the splitting of
the L energy level, but that effect is negligible for the elements listed here.)
The transition is from n = 2 to n = 1, so Eq. 40-26 combined with Eq. 40-24 yields
\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)
f=\left(\frac{m_e e^4}{8 \varepsilon_0^2 h^3}\right)\left(\frac{1}{1^2}-\frac{1}{2^2}\right)(Z-1)^2
so that the constant in Eq. 40-27 is
C=\sqrt{\frac{3 m_e e^4}{32 \varepsilon_0^2 h^3}}=4.9673 \times 10^7\, Hz ^{1 / 2}
using the values in the next-to-last column in the table in Appendix B (but note that the power of ten is given in the middle column).
We are asked to compare the results of Eq. 40-27 (squared, then multiplied by the accurate values of h/e found in Appendix B to convert to x-ray energies) with those in the table of K_\alpha energies (in eV) given at the end of the problem. We look up the corresponding atomic numbers in Appendix F.
(a) For Li, with Z = 3, we have
E_{\text {theory }}=\frac{h}{e} C^2(Z-1)^2=\frac{6.6260688 \times 10^{-34}\, J \cdot s }{1.6021765 \times 10^{-19}\, J / eV }\left(4.9673 \times 10^7\, Hz ^{1 / 2}\right)^2(3-1)^2=40.817 \,eV .
The percentage deviation is
\text { percentage deviation }=100\left(\frac{E_{\text {theory }}-E_{\exp }}{E_{\exp }}\right)=100\left(\frac{40.817-54.3}{54.3}\right)=-24.8 \% \approx-25 \%
In subsequent calculations, we use the steps outlined above.
(b) For Be, with Z = 4, the percentage deviation is –15%.
(c) For B, with Z = 5, the percentage deviation is –11%.
(d) For C, with Z = 6, the percentage deviation is –7.9%.
(e) For N, with Z = 7, the percentage deviation is –6.4%.
(f) For O, with Z = 8, the percentage deviation is –4.7%.
(g) For F, with Z = 9, the percentage deviation is –3.5%.
(h) For Ne, with Z = 10, the percentage deviation is –2.6%.
(i) For Na, with Z = 11, the percentage deviation is –2.0%.
(j) For Mg, with Z = 12, the percentage deviation is –1.5%.
Note that the trend is clear from the list given above: the agreement between theory and experiment becomes better as Z increases. One might argue that the most questionable step in Section 40-10 is the replacement e^4 \rightarrow (Z -1)^2 e^4 and ask why this could not equally well be e^4 \rightarrow ( Z-.9 ) ^2 e^4 \text { or } e^4 \rightarrow(Z-.8)^2 e^4 . For large Z, these subtleties would not matter so much as they do for small Z, since Z – ξ ≈ Z for Z >> ξ .