Question 6.8.Q1: Most of collision and radiation losses as a charged particle......

Most of collision and radiation losses as a charged particle penetrates into an absorber transfer in individual interactions only a minute fraction of the CP kinetic energy E_K to orbital electrons of the absorber. It is therefore convenient, as proposed by Berger and Seltzer, in 1983 to think of the CP moving through an absorber as if it is losing its kinetic energy gradually and continuously in a process called continuous slowing down approximation (CSDA). The CSDA range R_{CSDA} is then defined as

R_{\mathrm{CSDA}}=\int_0^{\left(E_{\mathrm{K}}\right)_0} \frac{\mathrm{d} E_{\mathrm{K}}}{S_{\mathrm{tot}}\left(E_{\mathrm{K}}\right)}          (6.109)

and can be determined through integrating the reciprocal of total mass stopping power \left(1/S_{tot}\right) over kinetic energy E_K from 0 to initial kinetic energy \left(E_K\right)_0. For heavy CPs, R_{CSDA} is a very good approximation to the mean range \bar{R} of the CP in the absorbing medium, because of the essentially rectilinear path of the CP in the absorbing medium.

(a) Use data available from the NIST to calculate the CSDA range R_{CSDA} in water for protons with incident kinetic energy \left(E_K\right)_0 of 100 MeV.
(b) Using data calculated in (a) plot RCSDA for protons in water against incident kinetic energy \left(E_K\right)_0 in the range from 0 ≤ \left(E_K\right)_0 ≤ 100 MeV.

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(a) In principle, to determine the CSDA range in water of protons with incident kinetic energy \left(E_K\right)_0 = 100 MeV we would use (6.109). However, since the reciprocal of the total mass stopping power S_{tot} is not available in an analytical form, we will resort to a numerical integration of 1/S_{tot} data and use a reasonably small energy interval \Delta E_K of 5 MeV in the integration. The steps in the numerical integration were discussed in Prob. 141, so here we present only a summary of the six steps involved:

(1) Decide on kinetic energy interval \Delta E_K for the numerical integration. We will use 5 MeV for \Delta E_K and thus need data for E_K in MeV of 0, 5, 10, 15, . . . , 95, 100.

(2) Obtain S_{col} in MeV · cm²/g (http://physics.nist.gov/PhysRefData/Star/Text/ PSTAR.html) from the NIST for protons in water for kinetic energies given in step (1). S_{tot} for heavy CP in general includes two terms: the predominant collision term S_{col} and a minor nuclear term S_{nuc} arising from elastic collisions between CP and the nucleus of the absorber. In our calculation we ignore the S_{nuc} term and assume that S_{tot} ≈ S_{col}. Note: for heavy CP radiation stopping power S_{rad} is zero in contrast to the situation with light CPs (electrons and positrons) for which S_{tot} = S_{col} + S_{rad}.

(3) Calculate 1/S_{col}, the reciprocal of S_{col}, for E_K given in step (1). Figure 6.24 plots 1/S_{col} against E_K for protons in water and also shows the energy intervals \Delta E_K used in the numerical integration

(4) Calculate 1/S_{col}, the average of 1/S_{col}, for each energy interval \Delta E_K. For example, for the last energy interval \Delta E_K shown in Fig. 6.24 we calculate 1/S_{col} as follows

\begin{aligned} \overline{1 / S_{\mathrm{col}}}(100 \mathrm{MeV})= & 1 / S_{\mathrm{col}}(100 \mathrm{MeV})-1 / S_{\mathrm{col}}(95 \mathrm{MeV}) \\ = & 0.5 \times\left[0.1372\left(\mathrm{MeV} \cdot \mathrm{cm}^2 / \mathrm{g}\right)^{-1}\right. \\ & \left.-0.1320\left(\mathrm{MeV} \cdot \mathrm{cm}^2 / \mathrm{g}\right)^{-1}\right] \\ = & 0.1346\left(\mathrm{MeV} \cdot \mathrm{cm}^2 / \mathrm{g}\right)^{-1}\quad (6.110) \end{aligned}(5) Calculate the area for each energy interval, i.e., \left(\overline{1 / S_{\mathrm{col}}}\right)_i \times \Delta E_{\mathrm{K}}.

(6) Sum up the areas \left(\overline{1 / S_{\mathrm{col}}}\right)_i \times \Delta E_{\mathrm{K}} \text { from area } 1 \text { to area } n \text { at }\left(E_{\mathrm{K}}\right)_0 to get

\sum_{i=1}^n\left(\overline{1 / S_{\mathrm{col}}}\right)_i \times \Delta E_{\mathrm{K}}=R_{\mathrm{CSDA}}\left[\left(E_{\mathrm{K}}\right)_0\right]        (6.111)

Table 6.18 summarizes the six steps in the calculation of the CSDA range RCSDA for 100 MeV protons in water and results in R_{CSDA} = 7.71 g/cm² = 7.71cm. This result, despite a relatively large energy interval of \Delta E_K = 5 MeV agrees well with the tabulated value of R_{CSDA} = 7.718 g/cm² obtained from the NIST.

(b) Data required for plotting of R_{CSDA} for protons in water in the kinetic energy interval 0 ≤ \left(E_K\right)_0 ≤ 100 MeV are actually available from row (6) of Table 6.18 where they were used to determine through numerical integration R_{CSDA} for 100 MeV protons. These data obviously contain range information for all proton energies up to \left(E_K\right)_0 = 100 MeV and all we need to do is extract it from the table. For example, for \left(E_K\right)_0 = 75 MeV we get directly from column (15) and row (6) a CSDA range R_{CSDA} of 4.61 cm; for \left(E_K\right)_0 = 20 MeV we get directly from column (4) and row (6) a CSDA range of 0.42 cm, etc.
In Fig. 6.25 we plot the CSDA range R_{CSDA} in water of protons with kinetic energy between 0 and 125 MeV. The data points represent calculated values listed in row (6) of Table 6.18 and the solid curve represents R_{CSDA} obtained from the NIST. The agreement between our data calculated with numerical integration using a relatively large energy interval of 5 MeV and data obtained from the NIST is excellent.

Table 6.18 Summary of data for calculation of CSDA range in water for protons of kinetic energy of 100 MeV. (*) indicates data obtained from the NIST
\begin{array}{lllllllllll} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline 1 E_{\mathrm{K}}(\mathrm{MeV}) & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50 \\ \hline 2 S_{\text {tot }}(*) & 79.11 & 45.67 & 32.92 & 26.07 & 21.75 & 18.76 & 16.56 & 14.88 & 13.54 & 12.45 \\ \hline 3 & 0.0126 & 0.0219 & 0.0304 & 0.0384 & 0.0460 & 0.0533 & 0.0604 & 0.0672 & 0.0739 & 0.0803 \\ \hline 4 \overline{1 / S_{\text {tot }}} & 0.0063 & 0.0173 & 0.0261 & 0.0344 & 0.0422 & 0.0500 & 0.0568 & 0.0638 & 0.0705 & 0.0771 \\ \hline 5\left(1 / S_{\text {tot }}\right) \Delta E_{\mathrm{K}} & 0.0316 & 0.0863 & 0.1307 & 0.1718 & 0.2108 & 0.2482 & 0.2842 & 0.3190 & 0.3527 & 0.3854 \\ \hline 6 \sum\left(1 / S_{\text {tot }}\right) \Delta E_{\mathrm{K}} & 0.0316 & 0.1179 & 0.2486 & 0.4205 & 0.6313 & 0.8795 & 1.1637 & 1.4827 & 1.8353 & 2.2208 \\ \hline \end{array}

\begin{array}{lllllllllll} \hline n & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\ \hline 1 E_{\mathrm{K}}(\mathrm{MeV}) & 55 & 60 & 65 & 70 & 75 & 80 & 85 & 90 & 95 & 100 \\ \hline 2 S_{\text {tot }}(*) & 11.54 & 10.78 & 10.13 & 9.559 & 9.063 & 8.625 & 8.236 & 7.888 & 7.573 & 7.289 \\ \hline 31 / S_{\text {tot }} & 0.0867 & 0.0928 & 0.0987 & 0.1046 & 0.1103 & 0.1159 & 0.1214 & 0.1268 & 0.1320 & 0.1372 \\ \hline 4 \overline{1 / S_{\text {tot }}} & 0.0835 & 0.0897 & 0.0957 & 0.1017 & 0.1075 & 0.1131 & 0.1187 & 0.1241 & 0.1294 & 0.1346 \\ \hline 5\left(1 / S_{\text {tot }}\right) \Delta E_{\mathrm{K}} & 0.4174 & 0.4486 & 0.4787 & 0.5083 & 0.5374 & 0.5657 & 0.5934 & 0.6205 & 0.6471 & 0.6731 \\ \hline 6 \sum\left(1 / S_{\text {tot }}\right) \Delta E_{\mathrm{K}} & 2.6382 & 3.0868 & 3.5655 & 4.0738 & 4.6112 & 5.1769 & 5.7703 & 6.3908 & 7.0378 & 7.7109 \\ \hline \end{array} \begin{array}{lllllllllll} \hline n & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 \\ \hline 1 E_{\mathrm{K}}(\mathrm{MeV}) & 105 & 110 & 115 & 120 & 125 & 130 & 135 & 140 & 145 & 150 \\ \hline 2 S_{\text {tot }}(*) & 7.030 & 6.794 & 6.577 & 6.377 & 6.192 & 6.021 & 5.861 & 5.713 & 5.575 & 5.445 \\ \hline 31 / S_{\text {tot }} & 0.1422 & 0.1472 & 0.1520 & 0.1568 & 0.1615 & 0.1661 & 0.1706 & 0.1750 & 0.1794 & 0.1837 \\ \hline 4 \overline{1 / S_{\text {tot }}} & & & & & & & & & & \\ \hline 5\left(1 / S_{\text {tot }}\right) \Delta E_{\mathrm{K}} & & & & & & & & & & \\ \hline 6 \sum\left(1 / S_{\text {tot }}\right) \Delta E_{\mathrm{K}} & & & & & & & & & & \\ \hline \end{array}
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