Question 6.4.Q5: Bethe equation for mass collision stopping power Scol of a s......

Bethe equation for mass collision stopping power S_{col} of a stopping medium (absorber) for heavy charged particles (CP) is given as

S_{\mathrm{col}}=4 \pi N_{\mathrm{e}}\left(\frac{e^2}{4 \pi \varepsilon_0}\right)^2 \frac{z^2}{m_{\mathrm{e}} c^2 \beta^2}\left\{\ln \frac{2 m_{\mathrm{e}} c^2}{I}+\ln \frac{\beta^2}{1-\beta^2}-\beta^2\right\}=C_1 \frac{N_{\mathrm{e}} z^2}{\beta^2} B_{\mathrm{col}},     (6.49)

with parameters defined in Prob. 131 and the Fano shell and density corrections neglected.

(a) Determine the mass collision stopping power S_{col} of water for a proton with incident kinetic energy E_K = 51 MeV. Mean ionization/excitation potential I of water is 75 eV; electron density N_e = ZN_A/A was determined in Prob. 133 as N_{e} = 3.343\times 10^{23} el/g.

(b) Determine the incident kinetic energy E_K of a deuteron for which water has the same mass collision stopping power S_{col} as for the proton in (a).

(c) Determine the incident kinetic energy E_K and mass collision stopping power S_{col} of water for the following particles: α particle, carbon-6 ion, and neon-10 ion having the same incident velocity β = υ/c as the proton in (a) and deuteron in (b).

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(a) To use (6.49) for calculation of S_{col} of water we will need the following parameters: (1) constant C_1, (2) velocity β = υ/c of proton with E_K = 51 MeV, and (3) collision stopping number of water for 51 MeV proton.

(1) From (6.18) in Prob. 131 we get the following expression for constant C_1.

\begin{aligned} C_1=4 \pi\left(\frac{e^2}{4 \pi \varepsilon_0}\right)^2 \frac{1}{m_{\mathrm{e}} c^2} & =4 \pi r_{\mathrm{e}}^2 m_{\mathrm{e}} c^2=4 \pi \times\left(2.818 \times 10^{-13}  \mathrm{~cm}\right)^2 \times(0.511  \mathrm{MeV}) \\ & =5.099 \times 10^{-25}  \mathrm{MeV} \cdot \mathrm{cm}^2\quad (6.50) \end{aligned} \begin{aligned} C_1=4 \pi\left(\frac{e^2}{4 \pi \varepsilon_0}\right)^2 \frac{1}{m_{\mathrm{e}} c^2} & =4π\left[\left(\frac{e^2}{4πε_0}\right)\frac{1}{(m_ec^2)}\right]^2m_ec^2=4 \pi r_{\mathrm{e}}^2 m_{\mathrm{e}} c^2=4 \pi \times\left(2.818 \times 10^{-13} \mathrm{~cm}\right)^2 \times(0.511  \mathrm{MeV}) \\ & =5.099 \times 10^{-25}  \mathrm{MeV} \cdot \mathrm{cm}^2\quad (6.18) \end{aligned}

(2) Velocity of the 51-MeV-proton is determined from the standard expression for relativistic kinetic energy E_K as follows

E_{\mathrm{K}}=(\gamma-1) E_0=\left(\frac{1}{\sqrt{1-\beta^2}}-1\right) m_{\mathrm{p}} c^2,            (6.51)

from where it follows that

\beta^2=1-\frac{1}{\left(1+\frac{E_{\mathrm{K}}}{m_{\mathrm{p}} c^2}\right)^2}=1-\frac{1}{\left(1+\frac{51}{938.3}\right)^2}=0.10 \quad \text { and } \quad \beta=0.01          (6.52)

(3) Atomic stopping number B_{col} of proton in water is determined as follows

\begin{aligned} B_{\mathrm{col}} & =\left\{\ln \frac{2 m_{\mathrm{e}} c^2}{I}+\ln \frac{\beta^2}{1-\beta^2}-\beta^2\right\} \\ & =\ln \frac{2 \times 0.511 \times 10^6  \mathrm{eV}}{75  \mathrm{eV}}+\ln \frac{0.1}{0.9}-0.1=9.52-2.197-0.10=7.22 .\quad (6.53) \end{aligned}

Now that we have all the required parameters we can determine S_{col} as follows

\begin{aligned} S_{\mathrm{col}} & =C_1 \frac{N_{\mathrm{e}} z^2}{\beta^2} B_{\mathrm{col}}=\left(5.099 \times 10^{-25} \mathrm{MeV} \cdot \mathrm{cm}^2\right) \times \frac{3.343 \times 10^{23}}{0.1 \mathrm{~g}} \times 7.22 \\ & =12.31 \frac{\mathrm{MeV} \cdot \mathrm{cm}^2}{\mathrm{~g}}\quad (6.54) \end{aligned}

and get a result that is in excellent agreement with the NIST mass collision stopping power S_{col} of 12.25 MeV · cm²/g obtained for a 51 MeV proton.

(b) For a heavy CP to engender in absorber the same mass collision stopping power S_{col} as does a proton, it must have the same velocity β as well as the same charge number z = 1. Since for both proton and deuteron z = 1, for a deuteron to engender S_{col} of 12.31 MeV · cm²/g in water, it must have the same velocity β of 0.01 as does a 51 MeV proton. Since the deuteron is heavier than the proton, its kinetic energy E_K(d) must be higher than that of the proton E_K(p). The ratio E_K(d)/E_K(p) is determined from (6.51) as follows

E_{\mathrm{K}}(\mathrm{p})=m_{\mathrm{p}} c^2\left(\frac{1}{\sqrt{1-\beta^2}}-1\right) \quad \text { and } \quad E_{\mathrm{K}}(\mathrm{d})=m_{\mathrm{d}} c^2\left(\frac{1}{\sqrt{1-\beta^2}}-1\right),               (6.55)

from where it follows that the ratio E_K(d)/E_K(p) is given as follows

\frac{E_{\mathrm{K}}(\mathrm{d})}{E_{\mathrm{K}}(\mathrm{p})}=\frac{m_{\mathrm{d}} c^2}{m_{\mathrm{p}} c^2}=\frac{1875.6}{938.3} \approx 2             (6.56)

Thus, for a deuteron to engender the same stopping power in water as does a proton, it must have the same velocity as the proton, and this implies that it has a kinetic energy that is twice the kinetic energy of the proton. Mass collision stopping power S_{col} of water is therefore 12.31 MeV · cm²/g for 51 MeV proton as well as for 102 MeV deuteron.

(c) We now calculate: (1) kinetic energy E_K(m_0) of particle with rest mass m_0 and (2) mass collision stopping power S_{col} of water for various CPs (deuteron, α particle, carbon-6 ion, and neon-10 ion), all of velocity β = 0.01, as determined for 51 MeV proton in (a).

(1) Kinetic energy E_K(m_0) of heavy CPs that all have the same velocity β is linearly proportional to CP’s rest energy E_0 = m_0c^2. Based on (6.56) we reach a general conclusion that kinetic energy of a given heavy CP of rest mass m_0 can be expressed in terms of kinetic energy of proton of rest mass m_p as

E_{\mathrm{K}}\left(m_0\right)=\frac{m_0 c^2}{m_{\mathrm{p}} c^2} E_{\mathrm{K}}\left(m_{\mathrm{p}}\right),              (6.57)

provided that both CPs have the same velocity β. Kinetic energy E_K(m_0) for proton, deuteron, α particle, carbon-6 ion, and neon-10 ion, all traveling with velocity β = 0.01, determined from (6.57) is listed in row (4) of Table 6.8, while kinetic energy of the various CPs stated in MeV/u is listed in row (5) of the table. It is evident that CPs with same MeV/u have the same velocity β and engender the same stopping number B_{col} in a given stopping material.

(2) Mass collision stopping power S_{col}(m_0,z) for the various CPs of rest mass m_0, all with the same velocity β = 0.01, will now be expressed in terms of S_{col}(m_p,z) of water for a proton. From (6.49) it is evident that, for a constant velocity β, S_{col}(m_0,z) is linearly proportional with z^2, where z is the number of charges that the CP carries, i.e.,

S_{\mathrm{col}}\left(m_0, z\right)=z^2 S_{\mathrm{col}}\left(m_{\mathrm{p}}, z\right)               (6.58)

Results for S_{col}(m_0,z) of water for proton, deuteron, α particle, carbon-6 ion, and neon-10 ion are summarized in row (7) of Table 6.8. Note that our S_{col} of water for α particles of E_K = 202.5 MeV is in excellent agreement with the NIST value of S_{col} = 49.03 MeV · cm²/g for the same conditions.

Table 6.8 Various parameters in calculation of kinetic energy E_K and mass stopping power of water for various heavy CPs (proton, deuteron, α particle, carbon-6 ion, and neon-10 ion)
1 Particle Proton Deuteron α particle Carbon-6 Neon-10
2 E_0=m_0 c^2 938.3 1875.6 3727.3 11174.9 18617.7
3 m_0 c^2 / m_{\mathrm{p}} c^2 1.0 2 3.97 11.91 19.84
4 E_{\mathrm{K}}\left(m_0\right) 51 102 202.5 11175 18616
5 E_{\mathrm{K}} / A(\mathrm{MeV} / \mathrm{u}) 51 ∼51 ∼51 ∼51 ∼51
6 z 1 1 2 6 10
7 S_{\mathrm{col}}\left(m_0, z\right)\left(\mathrm{MeV} \cdot \mathrm{cm}^2 / \mathrm{g}\right) 12.31 12.31 49.2 443.2 1231

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