Question 6.4.Q7: The Bethe equation for the mass collision stopping power Sco......

(a) To evaluate the link between (6.71) and (6.70) we compare the two equations directly and first determine the collision stopping power constant C_1 [see (6.18) in Prob. 131]

and then determine the functional form of the velocity function f (β) by rewriting (6.70) in a more convenient form as

S_{\mathrm{col}}=C_1 \frac{z^2 N_{\mathrm{e}}}{\beta^2}\left\{\ln \frac{2 m_{\mathrm{e}} c^2 \beta^2}{1-\beta^2}-\beta^2-\ln I\right\}           (6.73)

which allows us to define f (β) as follows

f(\beta)=\ln \frac{2 m_{\mathrm{e}} c^2 \beta^2}{1-\beta^2}-\beta^2=\ln \frac{1.022 \times 10^6 \beta^2}{1-\beta^2}-\beta^2            (6.74)

and also express the mass collision stopping power S_{col}, as shown in (6.71), where

C_1 is the collision stopping power constant given in (6.72).
f (β) is the velocity function given in (6.74).
ln I is the natural logarithm of the mean ionization/excitation potential of the absorber expressed in eV.

(b) The velocity function f (β) against kinetic energy E_K of the heavy CP is determined by first calculating β for a given E_K and then inserting the calculated β into (6.74) to calculate f (β). The results of the f (β) calculation for E_K in MeV = 0.01, 0.1, 1, 10, 100, and 1000 for protons are listed in Table 6.10 and plotted in Fig. 6.9 against normalized velocity β and in Fig. 6.10 against kinetic energy E_K.

(c) As evident from (6.71), to determine the mass collision stopping power S_{col} of water for a proton with kinetic energy E_K = 50 MeV we need to know the electron density N_e of water, mean ionization/excitation potential I of water, and the velocity β of a 50 MeV proton. N_e of water has been determined in Prob. 133 as N_e(water) = 3.343\times 10^{23} electron/g. I of water has been determined in Prob. 132 as I(water) = 74.4 eV.

Normalized velocity β of a 50 MeV proton is calculated from the standard expression that follows from the basic definition for relativistic kinetic energy E_{\mathrm{K}}=\left[1 /\left(1-\beta^2\right)^{1 / 2}-1\right] m_0 c^2, i.e.,

\beta^2=1-\frac{1}{\left(1+\frac{E_K}{m_{\mathrm{p}} c^2}\right)^2}=1-\frac{1}{\left(1+\frac{50}{938.3}\right)^2}=0.099 \quad \text { or } \quad \beta=0.314          (6.75)

Velocity function f (β) given in (6.74) yields the following value for a 50 MeV proton

f(\beta)=\ln \frac{1.022 \times 10^6 \beta^2}{1-\beta^2}-\beta^2=\ln \frac{1.022 \times 10^6 \times 0.099}{0.901}-0.099=11.53.             (6.76)

The mass collision stopping power S_{col} of water is thus given as

in excellent agreement with the value of 12.45 MeV · cm²/g provided by the NIST for the collision stopping power of water for 50 MeV proton.

(d) To calculate S_{col} of copper absorber traversed by a 250 MeV α particle using (6.71) we first determine β² for the α particle and then calculate the velocity function f (β) for the calculated β². We then insert f (β) into (6.71) to determine S_{col}.

Calculation of α particle velocity β from relativistic kinetic energy E_K.

\beta^2=1-\frac{1}{\left(1+\frac{E_{\mathrm{K}}}{m_{\mathrm{p}} c^2}\right)^2}=1-\frac{1}{\left(1+\frac{250}{3727.3}\right)^2}=0.122 \quad \text { or } \quad \beta=0.349          (6.78)

Calculation of velocity function f (β) using (6.74)

f(\beta)=\ln \frac{1.022 \times 10^6 \beta^2}{1-\beta^2}-\beta^2=\ln \frac{1.022 \times 10^6 \times 0.122}{1-0.122}-0.122=11.74          (6.79)

Calculation of mass collision stopping power S_{col} of copper for 250 MeV α particle

in good agreement with the NIST value of 27.24 MeV · cm²/g for mass collision stopping power of copper traversed by 250 MeV α particle.