Mean Excitation Energy

(1)

Amersham, Buckinghamshire, UK

11.1 Introductory Comments

11.2 Theoretical Evaluations of $\bar{I}$

11.2.1 Introductory Comments

11.2.2 Ratio of Bohr Velocity to Bohr Radius

11.2.3 Thomas–Fermi Model

11.2.4 Optical Dipole Oscillator Strength

11.2.5 Free Electron/Local Plasma Model

11.2.6 Condensed Media

11.3 Experimental Determination of $\bar{I}$

11.4 Parametric Representation of $\bar{I}$

Bibliography and Further Reading

Abstract

The mean excitation energy, $\overline{I}$ , appears in the logarithmic term of the soft collision stopping power. In practice, it is determined through extraction from measurements of the energy loss of a charged particle traversing a medium. However, there remains significant interest in understanding the theoretical aspects of the parameter, including deriving it ab initio and discovering that quite simplistic models can recreate the fundamental physical properties of $\overline{I}$ . In this chapter, theoretical models of increasing complexity are considered in the calculation of the mean excitation energy. The chapter is concluded with a review of the mechanisms of experimental determination of $\overline{I}$ and parameterisations of the quantity. Although the latter are of limited practical use since tabulated values of $\overline{I}$ exist, it is of historical interest to consider them.

11.1 Introductory Comments

The mean logarithmic excitation energy is defined by, following our discussion of Chap. 8,

$\ln \overline{I}=\sum\limits_{j=0}^f \ {{f_{j0 }}\ln \left( {{E_j}-{E_0}} \right)} .$

(11.1)

$\ln \overline{I}$ is a property of the medium alone and is independent of the nature of the charged projectile. Further, as $\bar{I}$ resides within the argument of a logarithm, changes and uncertainties in its value can have limited effects upon the value of the collision stopping power.

Values of $\overline{I}$ can be determined through both theoretical or experimental means. Because of the complexities inherent to the calculation and the need of knowledge of the atomic wavefunction, ab initio calculations of $\bar{I}$ have tended to be limited to monatomic gases of low atomic number. As such, the number of elements for which $\bar{I}$ has been calculated is small with the result that $\bar{I}$ is obtained largely through measurement. Both theoretical and experimental methods of evaluating $\bar{I}$ are investigated in this chapter.

We begin our discussion of the mean excitation energy with the theoretical means of determining the quantity. This is done, at first, by a very simple (and crude) estimate based upon the ratio of the Bohr velocity and atomic radius. Despite its crudeness, the estimate yields a surprisingly reasonable result. Then, following the work of Bloch (1933), we derive a more elegant means of calculating $\bar{I}$ using the Thomas–Fermi model of the atom that we derived in Chap. 5. We also examine the optical dipole oscillator strength and the local plasma approximation as approaches to calculating $\bar{I}$ .

Following this, we look at the considerations that must be accounted for in the experimental determination of $\bar{I}$ and the means of how it is measured. We then consider parameterisations of $\bar{I}$ as a function of Z which can be useful in practical medical radiation physics calculations instead of using tabular data.

Additional discussions of the mean excitation energy can be found in those reviews by Uehling (1954) and Fano (1964) and in ICRU Reports 37 (1984) and 49 (1993).

The general characteristics of $\bar{I}$ can be seen in Fig. 11.1 in which the mean excitation energy normalised to the atomic number for elements is shown as a function of atomic number. These data are from ICRU Report 37 and are largely based upon the $\bar{I}$ values obtained from energy loss transmission measurements of high-energy protons, as described in Sect. 11.3. It can be seen that this ratio decreases rapidly to a constant value of about 10 eV for Z greater than about 15. However, it is also seen that superimposed upon this general trend are discontinuities which can be attributed to the effects of atomic shell structure.

Fig. 11.1

Mean excitation energy normalised to the atomic number for elements as a function of atomic number (Data are from ICRU Report 37 (1984))

11.2 Theoretical Evaluations of $\bar{I}$

11.2.1 Introductory Comments

A theoretical evaluation of $\overline{I}$ will likely have limited application to medical radiation physics calculations. The desired accuracy in the numerical value of the quantity will be best provided by experiment, as discussed in Sect. 11.3. Even so, a theoretical understanding of $\overline{I}$ can be revelatory by showing facets of the underlying physics that plays a role in the features of the mean excitation energy. In this section, we review a number of theoretical means of calculating $\overline{I}$ ab initio.

11.2.2 Ratio of Bohr Velocity to Bohr Radius

We can estimate crudely the dependence of the excitation energy upon the atomic number of the atom by treating the atomic electrons as oscillators and by approximating the atomic electron orbital frequency with the ratio of the Bohr velocity to the Bohr radius:

$\begin{array}{cclcllclcl} \omega \approx & \frac{{{v_0}}}{{{r_{\infty }}}} \\\approx & \frac{{Z\alpha c}}{{\left( {\frac{{\hbar c}}{{\alpha {m_{\mathrm{ e}}}}}} \right)}} \\\approx & Z\frac{{{\alpha^2}{m_{\mathrm{ e}}}}}{\hbar }.\end{array}$

(11.2)

Then, the mean excitation energy is

$\begin{array}{cclcllclcl} \bar{I}\approx \ & \hbar \omega \\\approx & \ Z{\alpha^2}{m_{\mathrm{ e}}} \\\approx & \ 27Z\mathrm{ eV}.\end{array}$

(11.3)

There should be no doubt that this is a crude model. However, the result suggests that the ratio of the mean excitation energy of a monatomic medium to its atomic number, $\frac{\bar{I}}{Z}$ , is a constant, which was shown to be reasonably true for $$ Z>15 $$” src=”/wp-content/uploads/2016/04/A306762_1_En_11_Chapter_IEq001123.gif”></SPAN> in Fig. <SPAN class=InternalRef><A href=

$$ Z>15 $$” src=”/wp-content/uploads/2016/04/A306762_1_En_11_Chapter_IEq001123.gif”></SPAN> in Fig. <SPAN class=InternalRef><A href=

11.1. The constant of proportionality in (11.3) is to be expected to be an overestimate as we have used the Bohr velocity for the ground state, ${v_0}=Z\alpha c$ , which is the fastest of the atomic electrons. If we were to instead use the arithmetic mean and approximate a mean Bohr velocity, ${{\bar{v}}_0}\approx \frac{{Z\alpha c}}{2}$ , we obtain

$\bar{I}\approx 13.5Z\mathrm{ eV}.$

(11.4)

Quantitatively, this new result is closer to the results obtained from experiment as shown in Fig. 11.1.

We now attempt to develop further this ‘back-of-the-envelope’ calculation further by returning to the Thomas–Fermi statistical model of the atom that we developed in Chap. 5.¹

11.2.3 Thomas–Fermi Model

Another means of estimating $\bar{I}$ is through the statistical model of Thomas and Fermi in describing the atom. This model was derived in Sect. 5.3.5 in the context of describing the atomic electron screening of the nuclear Coulomb potential. Recall that the radius of the Thomas–Fermi atom is defined as

$\begin{array}{cclcllclcl} {r_{\mathrm{ TF}}}\equiv & \frac{{{{{\left( {3\pi } \right)}}^{{\frac{2}{3}}}}}}{{{2^{{\frac{7}{3}}}}}}\frac{{{r_{\infty }}}}{{{Z^{{\frac{1}{3}}}}}} \\\approx & 0.885\frac{{{r_{\infty }}}}{{{Z^{{\frac{1}{3}}}}}}.\end{array}$

We will crudely model the atom as being a homogeneous sphere of negative charge –Ze surrounding a point charge +Ze. The sphere, or electron ‘cloud’, will be subject to two sources of potential energy, one due to the self-energy of the electrons, which we refer to as U _e, and the other due to that of the Coulomb interaction with the nucleus, which is referred to as U _n. In addition, there will be a kinetic energy contribution. Let us now evaluate these three sources.

First, we evaluate the self-energy, U _e . Consider the electron sphere to have a radius r _TF for a net charge of –Ze. Because of the assumption of the negative charge homogeneity, the negative charge density within the sphere is

${\rho_{\mathrm{ e}}}=\frac{3Ze }{{4\pi r_{\mathrm{ TF}}^3}}.$

(11.5)

Then, the self-energy term is

${U_{\mathrm{ e}}}=\frac{{{\rho_{\mathrm{ e}}}}}{{3{\varepsilon_0}}}\int\limits_0^{{{r_{\mathrm{ TF}}}}} {{{\mathrm{ d}}^3}{\bf r}\ {r^2}} .$

(11.6)

The integration of (11.6) results in

$\begin{array}{cclcllclcl} {U_{\mathrm{ e}}}&= \frac{3}{5}\frac{{{{{\left( {Ze} \right)}}^2}}}{{4\pi {\varepsilon_0}{r_{\mathrm{ TF}}}}} \\&= \frac{3}{5}\alpha \hbar c\frac{{{Z^2}}}{{{r_{\mathrm{ TF}}}}}\end{array}$

(11.7)

where the definition of the fine-structure constant has been introduced in order to simplify the form of the expression.

The second term consists of the energy that is required to bring the nucleus from infinity to the edge of the electron cloud and then to the centre. We label these two individual terms as U _n1 and U _n2, respectively. The first is simply

$\begin{array}{cclcllclcl} {U_{\mathrm{ n}1}}&= -\frac{1}{{4\pi {\varepsilon_0}}}\frac{{{{{\left( {Ze} \right)}}^2}}}{{{r_{\mathrm{ TF}}}}} \\&= -\alpha \hbar c\frac{{{Z^2}}}{{{r_{\mathrm{ TF}}}}}.\end{array}$

(11.8)

To calculate ${U_{\mathrm{ n}2}}$ , we note that the electric field within the sphere is

$\begin{array}{cclcllclcl} \left| {\mathbf{E}(r)} \right|&= \frac{Ze }{{4\pi {\varepsilon_0}{r^2}}}\left( {{\rho_{\mathrm{ e}}}\frac{{4\pi }}{3}{r^3}} \right) \\&= \frac{{Ze{\rho_{\mathrm{ e}}}}}{{3{\varepsilon_0}}}r.\end{array}$

(11.9)

Then, the energy required to bring the nucleus to the centre of the electron cloud is

$\begin{array}{cclcllclcl} {U_{\mathrm{ n}2}}&= -\frac{1}{2}\left| {\mathbf{E}(r)} \right|r \\&= -\frac{1}{2}\alpha \hbar c\frac{{{Z^2}}}{{{r_{\mathrm{ TF}}}}}.\end{array}$

(11.10)

Thus, the potential energy of the electron cloud containing the nucleus is the sum of ${U_{\mathrm{ e}}}$ , ${U_{{\mathrm{ n}1}}}$ and ${U_{{\mathrm{ n}2}}}$ , or

${U_{\mathrm{ e}}}=-\frac{9}{10}\alpha \hbar c\frac{{{Z^2}}}{{{r_{\mathrm{ TF}}}}}.$

(11.11)

We now calculate the kinetic energy using the Fermi gas model (see Section 3.3 of McParland (2010) for a derivation of that model). The mean energy of the electron states is given by

${E_{\mathrm{ e}}}=\frac{3}{5}Z\frac{{p_{\mathrm{ F}}^2}}{{2{m_{\mathrm{ e}}}}}$