Determine the constant $\mathbf{C}$ in Eq. 40-27 to five significant figures by finding 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 in Eq. 40-27, determine the theoretical energy ${\mathbf{E}}_{\mathbf{theory}}$of the ${\mathbf{K}}_{\mathbf{\alpha }}$photon for the low-mass elements listed in the following table. The table includes the value (eV) of the measured energy ${\mathbf{E}}_{\mathbf{exp}}$of the ${\mathbf{K}}_{\mathbf{\alpha }}$photon for each listed element. The percentage deviation between ${\mathbf{E}}_{\mathbf{theory}}$and ${\mathbf{E}}_{\mathbf{exp}}$can be calculated as:

$\mathbf{percentage}\mathbf{}\mathbf{deviation}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{theory}}\mathbf{-}{\mathbf{E}}_{\mathbf{exp}}}{{\mathbf{E}}_{\mathbf{exp}}\mathbf{\times }\mathbf{100}}$

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?

(There is actually more than one ${\mathbf{K}}_{\mathbf{\alpha }}$ray because of the splitting of the energy level, but that effect is negligible for the elements listed here.)

The experimental energies of the elements is given is the table.

The given formula of percentage deviation is,

$\%\mathrm{deviation}=\frac{{\mathrm{E}}_{\mathrm{theory}}-{\mathrm{E}}_{\exp }}{{\mathrm{E}}_{\exp }\times 100}$

Using the Rydberg-Equation of energy for a transition of an electron from one state n=2 to n=1 considering equations 40-26, you can get the value of the constant C. Then, using this value in the calculated energy theory equation; you can get the theoretical energy of the different elements as per their atomic numbers. Now, the percentage deviation of each element can be calculated using the given experimental data and the equation given.

Formulae:

The energy equation of a hydrogen atom for $K_{\alpha }$transition line from equation (40.24),

$$\begin{gathered} E=-\left(\frac{m_e{e}^4}{8{\epsilon }_0^2{h}^2}\right)\left[\frac{1}{2^2}-\frac{1}{1^2}\right]{\left(Z-1\right)}^2 \\ =\left(\frac{3m_e{e}^4}{32{\epsilon }_0^2{h}^2}\right){\left(Z-1\right)}^2 \end{gathered}$$ ….. (i)

Here, $m_e$ is the mass of electron, e is the charge of electron, ${\epsilon }_0$ is the permittivity of free space in vacuum,h is Planck’s constant, and Z is the atomic number.

The theoretical energy referring to the Appendix B,

$E_{theory}=\frac{h}{e}{C}^2{\left(Z-1\right)}^2$ ….. (ii)

Here, C is the constant.

The energy of a photon from Planck’s constant,

E =hf ….. (iii)

Here, fis the frequency.

The frequency equation for the $K_{\alpha }$line from equation 40.27,

$\sqrt{f}=CZ-C$ ….. (iv)

Comparing equations (i) and (iii), you can get the frequency value as:

$f=\left(\frac{3m_e{e}^4}{32{\epsilon }_0^2{h}^2}\right){\left(Z-1\right)}^2$

Now using equation (iv) and above value, the constant C can be calculated as follows:

$$\begin{gathered} \sqrt{\left[\left(\frac{3m_e{e}^4}{32{\epsilon }_0^2{h}^2}\right){\left(Z-1\right)}^2\right]}=C\left(Z=1\right) \\ {C}^2{\left(Z-1\right)}^2=\left[\left(\frac{3m_e{e}^4}{32{\epsilon }_0^2{h}^2}\right){\left(Z-1\right)}^2\right] \\ {C}^2=\left(\frac{3m_e{e}^4}{32{\epsilon }_0^2{h}^2}\right) \end{gathered}$$

….. (v)

Now, using the values in the next-to-last column in the table in Appendix B, you can get the value of the constant as follows:

Mass of electron, ${\mathrm{m}}_{\mathrm{e}}=9.1\times {10}^{-31}\mathrm{kg}$,

The charge of electron, $e=1.6\times {10}^{-19}C$,

The permittivity of free space in vacuum, localid="1661925188907" ${\epsilon }_0=8.85\times {10}^{12}\frac{F}{m},$

$\mathrm{Planck}\text{'}\mathrm{sconstant},"\mathrm{h}=6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}$

Therefore, the value of constant C is,

$$\begin{gathered} \mathrm{C}=\sqrt{\frac{3\left(9.1\times {10}^{-31}\mathrm{kg}\right){\left(1.6\times {10}^{-19}\mathrm{C}\right)}^4}{32{\left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right)}^2{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^3}} \\ =\sqrt{\frac{178.91328\times {10}^{-107}}{729106.2387\times {10}^{-126}}} \\ =\sqrt{0.000245387\times {10}^{19}} \\ =4.9673\times {10}^7{\mathrm{Hz}}^{1/2} \end{gathered}$$

Now, the theoretical energy for element, Li (Z=3) can be calculated using the above C value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(3-1\right)}^2 \\ =40.817\mathrm{eV} \end{gathered}$$

Thus, the percentage deviation of energy for the element Li is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(40.817\mathrm{eV}-54.3\mathrm{eV}\right)}{54.3\mathrm{eV}}\times 100 \\ =-24.8\% \\ \approx -25\% \end{gathered}$$

Hence, the percentage deviation is -25%.

Now, the theoretical energy for element, Be (Z=4) can be calculated using the above C value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(4-1\right)}^2 \\ =91.838\mathrm{eV} \end{gathered}$$

Thus, the percentage deviation of energy for the element Be is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(91.838\mathrm{eV}-108.5\mathrm{eV}\right)}{108.5\mathrm{eV}}\times 100 \\ =-15.36\% \\ \approx -15\% \end{gathered}$$

Hence, the percentage deviation is -15%.

Now, the theoretical energy for element, B (Z=5) can be calculated using the above C value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(5-1\right)}^2 \\ =163.268\mathrm{eV} \end{gathered}$$

So, the percentage deviation of energy for the element B is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(163.268\mathrm{eV}-183.3\mathrm{eV}\right)}{183.3\mathrm{eV}}\times 100 \\ =-10.93\% \\ \approx -11\% \end{gathered}$$

Hence, the percentage deviation is -11%.

Now, the theoretical energy for element, C (Z=6) can be calculated using the above value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(6-1\right)}^2 \\ =255.106\mathrm{eV} \end{gathered}$$

Therefore, the percentage deviation of energy for the element C is given using the formula as follows:

role="math" localid="1661924438928" $$\begin{gathered} \%\mathrm{deviation}=\frac{\left(255.106\mathrm{eV}-277\mathrm{eV}\right)}{277\mathrm{eV}}\times 100 \\ =-7.9\% \end{gathered}$$

Hence, the percentage deviation is -7.9%.

Now, the theoretical energy for element, N (Z=7) can be calculated using the above value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(7-1\right)}^2 \\ =367.353\mathrm{eV} \end{gathered}$$

Hence, the percentage deviation of energy for the element N is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(367.353\mathrm{eV}-393.4\mathrm{eV}\right)}{392.4\mathrm{eV}}\times 100 \\ =-6.38\% \\ \approx -6.4\% \end{gathered}$$

Therefore, the percentage deviation is 6.4%.

Now, the theoretical energy for element, O (Z=8) can be calculated using the above C value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(8-1\right)}^2 \\ =500.008\mathrm{eV} \end{gathered}$$

Thus, the percentage deviation of energy for the element O is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(500.008\mathrm{eV}-524.9\mathrm{eV}\right)}{524.9\mathrm{eV}}\times 100 \\ =--4.74\% \\ \approx -4.7\% \end{gathered}$$

Hence, the percentage deviation is -4.7% .

Now, the theoretical energy for element, F (Z=9) can be calculated using the above F value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(9-1\right)}^2 \\ =653.072\mathrm{eV} \end{gathered}$$

So, the percentage deviation of energy for the element F is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(653.072\mathrm{eV}-676.8\mathrm{eV}\right)}{676.8\mathrm{eV}}\times 100 \\ =--3.51\% \\ \approx -3.5\% \end{gathered}$$

Hence, the percentage deviation is -3.5%.

Now, the theoretical energy for element, Ne (Z=10) can be calculated using the above C value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(10-1\right)}^2 \\ =826.544\mathrm{eV} \end{gathered}$$

Thus, the percentage deviation of energy for the element Ne is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(826.544\mathrm{eV}-848.6\mathrm{eV}\right)}{524.9\mathrm{eV}}\times 100 \\ =--2.599\% \\ \approx -2.6\% \end{gathered}$$

Hence, the percentage deviation is -2.6%.

Now, the theoretical energy for element, Na can be calculated using the above value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(11-1\right)}^2 \\ 1020.425\mathrm{eV} \end{gathered}$$

So, the percentage deviation of energy for the element Na is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(1020.425\mathrm{eV}-1041\mathrm{eV}\right)}{1041\mathrm{eV}}\times 100 \\ =--1.99\% \\ \approx -2\% \end{gathered}$$

Therefore, the percentage deviation is -2%.

Now, the theoretical energy for element, Mg (Z=12) can be calculated using the above C value in equation (ii) as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{theory}}=\frac{6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}}{\left(4.9673\times {10}^7{\mathrm{Hz}}^{1/2}\right)}^2{\left(12-1\right)}^2 \\ =1234.714eV \end{gathered}$$

So, the percentage deviation of energy for the element Mg is given using the formula as follows:

$$\begin{gathered} \%\mathrm{deviation}=\frac{\left(1234.714\mathrm{eV}-1254\mathrm{eV}\right)}{1254\mathrm{eV}}\times 100 \\ =--1.54\% \\ \approx -1.5\% \end{gathered}$$

Hence, the percentage deviation is 1.5%.