- #1
lavster
- 217
- 0
can someone confirm that the equation for the linear energy transfer is:
[tex]LET=-\frac{4\pi e^4Nz^2Z}{m_0v^2} (ln\frac{2m_0v^2}{I}-ln(1-\beta^2)-\beta^2)=-\frac{dE}{dx},[/tex]
where e is the charge of an electron, Z is the atomic number of the material being irradiated, m_0 is the mass, z is the charge number of the beam particle, v is the velocity of the beam particle, I is the average ionisation potential ([tex]\approx11.5Z(eV)[/tex]), [tex]\beta=\frac{v}{c}[/tex] and [tex]\frac{dE}{dx}[/tex] is the energy loss per unit length.
and hence LET is the same as the stopping power. (the internet as confused me greatly)
thanks
[tex]LET=-\frac{4\pi e^4Nz^2Z}{m_0v^2} (ln\frac{2m_0v^2}{I}-ln(1-\beta^2)-\beta^2)=-\frac{dE}{dx},[/tex]
where e is the charge of an electron, Z is the atomic number of the material being irradiated, m_0 is the mass, z is the charge number of the beam particle, v is the velocity of the beam particle, I is the average ionisation potential ([tex]\approx11.5Z(eV)[/tex]), [tex]\beta=\frac{v}{c}[/tex] and [tex]\frac{dE}{dx}[/tex] is the energy loss per unit length.
and hence LET is the same as the stopping power. (the internet as confused me greatly)
thanks