Question 6.5.Q2: Expressions for the mass collision stopping power Scol of a ......
Expressions for the mass collision stopping power S_{col} of a given absorber for electron and positron traversing the absorber are of the same form except for the difference in the stopping power functions F^{−} for electron and F^{+} for positron. The two functions are given as follows
F^{-}(\tau)=\frac{1}{(\tau+1)^2}\left[1+\frac{\tau^2}{8}-(2 \tau+1) \ln 2\right] (6.90)
and
F^{+}(\tau)=2 \ln 2-\frac{\tau(\tau+2)^2}{12(\tau+1)^2}\left[23+\frac{14}{\tau+2}+\frac{10}{(\tau+2)^2}+\frac{4}{(\tau+2)^3}\right], (6.91)
where τ is the kinetic energy of the electron and positron E_K normalized to the electron/positron rest energy m_{\mathrm{e}} c^2 \text {, i.e., } \tau=E_{\mathrm{K}} /\left(m_{\mathrm{e}} c^2\right).
(a) Figure 6.13 plots F^{-}(\tau) \text { and } F^{+}(\tau) \text { against kinetic energy } E_{\mathrm{K}}. Calculate F^{-}(\tau) \text { and } F^{+}(\tau) \text { for } E_{\mathrm{K}}=0.337 \mathrm{MeV} using (6.90) and (6.91), respectively, and verify that your results fit on the F^{ \pm} graph of Fig. 6.13.
(b) Determine analytically the following features of the two stopping power functions F^{-} \text {and } F^{+} \text {: (1) } \lim _{\tau \rightarrow 0} F^{-} \text {; (2) } \lim _{\tau \rightarrow 0} F^{+} \text {; (3) } \lim _{\tau \rightarrow \infty} F^{-}; (4) \lim _{\tau \rightarrow \infty} F^{+} \text {; (5) } \tau_0^{-} \text {for } F^{-}\left(\tau_0^{-}\right)=0 \text {; and (6) } \tau_*^{-} for absolute minimum in F^{−}.
(c) Use Fig. 6.13 to determine the following features of the stopping power function F^{+} \text {and } F^{+} \text {: (1) } \tau_0^{+} \text {for } F^{+}\left(\tau_0^0\right)=0 \text {; (2) } \tau^{\prime} \text { for } F^{+}\left(\tau^{\prime}\right)=F^{-}\left(\tau^{\prime}\right) ; \text { and (3) } \tau_*^{+} for absolute minimum in F^{+}.
(d) Comment on the behavior of the two stopping power functions F^{−}\ and\ F^{+} with respect to the kinetic energy E_K of the electron and positron, respectively.
(a) We first determine \tau=E_{\mathrm{K}} /\left(m_{\mathrm{e}} c^2\right)=0.337 / 0.511=0.66 for the light CP of kinetic energy E_K = 0.337 MeV and then insert τ into (6.90) and (6.91) to get
As shown in Fig. 6.13, the two functions F^{-} \text {and } F^{+} are identical and equal to −0.2 at 0.337 MeV.
(b) The stopping power functions for electron and positron are from (6.90) and (6.91) given as
\lim _{\tau \rightarrow 0} F^{-}=\lim _{\tau \rightarrow 0}\left\{\frac{1}{(\tau+1)^2}\left[1+\frac{\tau^2}{8}-(2 \tau+1) \ln 2\right]\right\}=1-\ln 2=0.3069 (6.94)
(2)
\begin{aligned} \lim _{\tau \rightarrow 0} F^{+}(\tau) & =\left\{2 \ln 2-\frac{\tau(\tau+2)}{12(\tau+1)^2}\left[23+\frac{14}{\tau+2}+\frac{10}{(\tau+2)^2}+\frac{4}{(\tau+2)^3}\right]\right\} \\ & =2 \ln 2 \approx 1.386 .\quad (6.95) \end{aligned}(3)
\lim _{\tau \rightarrow \infty} F^{-}=\lim _{\tau \rightarrow \infty}\left\{\frac{\frac{1}{\tau^2}+\frac{1}{8}-\frac{2 \ln 2}{\tau}+\frac{\ln 2}{\tau^2}}{1+\frac{2}{\tau}+\frac{1}{\tau^2}}\right\}=\frac{1}{8}=0.125 . (6.96)
(4)
\begin{aligned} \lim _{\tau \rightarrow \infty} F^{+} & =\lim _{\tau \rightarrow \infty}\left\{2 \ln 2-\frac{1+\frac{2}{\tau}}{12+\frac{24}{\tau}+\frac{12}{\tau^2}}\left[23+\frac{14}{\tau+2}+\frac{10}{(\tau+2)^2}+\frac{4}{(\tau+2)^3}\right]\right\} \\ & =2 \ln 2-\frac{23}{12}=-0.5304 .\quad (6.97) \end{aligned}(5)
\begin{aligned} & F^{-}\left(\tau_0^{-}\right) \frac{1+\frac{\left(\tau_0^{-}\right)^2}{8}-\left(2 \tau_0^{-}+1\right) \ln 2}{\left(\tau_0^{-}+1\right)^2}=0 \quad \text { or } \\ & \left(\tau_0^{-}\right)^2-(16 \ln 2) \tau-8(\ln 2-1)=0, \end{aligned} (6.98)
\begin{aligned} & \tau_0^{-}=\frac{16 \ln 2 \pm \sqrt{(16 \ln 2)^2+32(\ln 2-1)}}{2} \text { or } \\ & \left(\tau_0^{-}\right)_1=0.225 \rightarrow E_{\mathrm{K}}\left[\left(\tau_0^{-}\right)_1\right]=0.115 \mathrm{MeV} \\ & \left(\tau_0^{-}\right)_2=10.87 \rightarrow E_{\mathrm{K}}\left[\left(\tau_0^{-}\right)_2\right]=5.55 \mathrm{MeV} . \end{aligned} (6.99)
(6) To find the absolute minimum in F^{-} \text {we set } \mathrm{d} F^{-} /\left.\mathrm{d} \tau\right|_{\tau=\tau_*^{-}}=0 and get
\left.\frac{\mathrm{d} F^{-}}{\mathrm{d} \tau}\right|_{\tau=\tau_*^{-}}=-\frac{2}{(1+\tau)^3}\left[1+\frac{\tau^2}{8}-(2 \tau+1) \ln 2\right]+\frac{1}{(1+\tau)^2}\left[\frac{2 \tau}{8}-2 \ln 2\right]=0 (6.100)
resulting in
-2+\tau\left(2 \ln 2+\frac{1}{4}\right)=0 \quad \text { and } \quad \tau_*^{-}=1.222 \quad \text { or } \quad E_{\mathrm{K}}^*=\tau_*^{-} m_{\mathrm{e}} c^2=0.625 \mathrm{MeV} (6.101)
Results of calculations (1) through (6) are displayed in Fig. 6.13 as data points (1) through (6) on the F^{−}\ and\ F^{+} stopping power curves.
(c) We now determine \text { e } \tau_0^{+}, \tau^{\prime}, \text { and } \tau_*^{+} by reading the values directly from Fig. 6.13 and get: \tau_0^{+}=0.482 \text { with } E_{\mathrm{K}}\left(\tau_0^{+}\right)=\tau_0^{+} m_{\mathrm{e}} c^2=0.247 \mathrm{MeV} ; \tau^{\prime}=-0.66 with E_{\mathrm{K}}\left(\tau^{\prime}\right)=\tau^{\prime} m_{\mathrm{e}} c^2=0.337 \text {, and } \tau_*^{+}=3.52 \text { with } E_{\mathrm{K}}\left(\tau_*^{+}\right)=\tau_*^{+} m_{\mathrm{e}} c^2=1.8 \mathrm{MeV}.
(d) Mass collision stopping powers of a given material differ for electron and positron because of the differences in the two stopping power functions F^{−}\ and\ F^{+}, respectively, plotted in Fig. 6.13. At low kinetic energies E_{\mathrm{K}} \text {, function } F^{-}(\tau) exceeds function F^{+}(\tau) by a factor of 1.386/0.3069 ≈ 4.5, as determined in (b). Both F^{-} \text {and } F^{+} then decrease with increasing E_K, become negative crossing the abscissa axis at ∼0.115 MeV and ∼0.24 MeV, respectively, and become equal (−0.2) at ∼ 0.3 MeV. Both functions then go through their respective absolute minima at around E_K ≈ 1 MeV and then attain saturation at very high E_K;F^{+} saturates at ∼−0.53 and F^{−} at ∼0.125, as also determined in (b).