- #1
James_1978
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My question is taking the Bethe-Bloch equation and integrating to find the range of an energetic Deuteron. I first have the Bethe-Bloch equation,
[tex] \frac{dE}{dx} = -\rho 0.1535 \frac{Z_{p}^{2}}{\beta^{2}} \left(\frac{Z_{A}}{A}\right) \left[ln\left(\frac{2 M_{e} c^{2} \beta^{2}}{IE\left(1 - \beta^{2}\right)}\right)^{2} - 2\beta^{2}\right] [/tex]
To find the Range we integrate the following equation,
[tex] \mbox{Range} = \int_{E}^{0} \frac{1}{dE/dx}dE [/tex]
I have tried to integrate this equation but I am unable to get it to work. My integral looks like the following,
[tex] \mbox{Range} = \int_{E}^{0} \frac{1}{\frac{1}{E}\left(ln\left[\frac{E}{\left(1-E\right)}\right]^{2} - 2*E\right)}dE [/tex]
Where [itex] \beta [/itex] is a function of Energy. What am I doing incorrect?
TIA
[tex] \frac{dE}{dx} = -\rho 0.1535 \frac{Z_{p}^{2}}{\beta^{2}} \left(\frac{Z_{A}}{A}\right) \left[ln\left(\frac{2 M_{e} c^{2} \beta^{2}}{IE\left(1 - \beta^{2}\right)}\right)^{2} - 2\beta^{2}\right] [/tex]
To find the Range we integrate the following equation,
[tex] \mbox{Range} = \int_{E}^{0} \frac{1}{dE/dx}dE [/tex]
I have tried to integrate this equation but I am unable to get it to work. My integral looks like the following,
[tex] \mbox{Range} = \int_{E}^{0} \frac{1}{\frac{1}{E}\left(ln\left[\frac{E}{\left(1-E\right)}\right]^{2} - 2*E\right)}dE [/tex]
Where [itex] \beta [/itex] is a function of Energy. What am I doing incorrect?
TIA
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