Linear energy transfert (LET)

1. Definition

Radiations interact with target matter, mainly by ionization process. They lose their energy through interactions with the atoms. The average amount of energy that is lost over a defined distance, is known as the Linear Energy Transfer (LET) in radiation biology, and stopping power in nuclear physics. In other words, LET is the energy loss per unit length of particle track (passage through a medium).In this section we are going to consider high energy heavy particles, so they transfer a small fraction of their during a single electronic collision, therefore, their deflection in the collision is negligible, and their travel is essentially a straight path in irradiated matter.

Heavy charged particles are "high LET radiation" while x-rays, wheras gamma-rays, and fast electrons are "low LET radiation". Using relativistic quantum mechanics, Bethe derived the an expression for the LET of a uniform target medium for a heavy charged particle.

2. Bethe's formula

The relativistic Bethe's formula is:
- dE/dx = (4πk_e²z²e⁴N/mc²β²) {ln[(2mc²β²)/(I(1 - β²))] - β²}
Where:
k_e = 8.99 x 10⁹ N m² C^-2, Coulomb's constant k_e = 1/4πε₀
Permittivity constant: ε₀ = 8.85 x 10^-12 C²/N.m²
z = atomic number of the incident heavy particle,
e = charge of the electron = 1,6 x 10^{- 19} C,
N = number of electrons per unit volume in the medium,
m = electron rest mass = 9.11 x 10 ^-31 kg, and mc² = 0.511 MeV,
c = speed of light in vacuum = 3.00 x 10⁸ m/s,
� = V/c = initial speed of the particle relative to c,
I = mean excitation energy of the medium.

3. Mean excitation energy

The mean excitation energy is measured in experiments. We have the following empirical formulas to estimate the values of I for medium of charge element Z, in eV:

4. Reduced Bethe's formula

Per volume unit = 1 m³ , and with:
1 eV = 1.602 � 10^-19 joules, 1 MeV = 1.602 � 10^-13 joules, and mc² = 0.511 MeV, we have:
4πk_e²e⁴/mc² = 5.085 x 10 ^{- 32} keV/µm.
If we introduce the function ζ such as:
ζ(β) = ln[(2mc²β²)/(1 - β²)] - β²
= ln[(1.02 x 10⁶ β²)/(1 - β²)] - β²,
the Bethe's furmula takes the reduced following form:
- dE/dx = (5.085 x 10 ^{- 32} z²N/β²) [ζ(β) - ln I] keV/µm.

- dE/dx = (5.085 x 10 ^{- 32} z²N/β²) [ζ(β) - ln I] keV/µm.
ζ(β) = ln[(1.02 x 10⁶ β²)/(1 - β²)] - β²

Recall I is in units of eV. The function ζ(β) depends only on the velocity of the projectile ( incident radiation).
We use also the Mass Stopping Power quantity, which is the LET divided by the density of the medium.
Remark:
The Bethe's formula remains an approximation for the real facts. When approching β to zero, the LET becomes infinite!. The β in the expression of the LET is related to the initial velocity; then a constant!.

5. Reduced Bethe's formula for water target

For water target:
Molecular weight = 18.0 g/mol; the number of electrons per molecule is 10.
One m³ weights of 10⁶ g. The density of electrons N is:
N = 10 x 6.023 x 10²³ electrons/mol = 10 x 6.023 x 10²³/18.0 x 10^-6 electrons/m³ N = 3.35 x 10²⁹ electrons/m³

With I_ex ≈ 74.6 eV, we have:
- dE/dx = (5.085 x 10 ^{- 32} z² x 3.35 x 10²⁹ /β²) {9.525 + ln[β²/(1 - β²)] - β²} keV/µm.
= (0.0170 x z²) {(9.525/β²) + (1/β²)ln[β²/(1 - β²)] - 1} keV/µm.

We have:
1/β² = c²/v²
If the initial energy of the projectile is E = (1/2)mv², then v² = 2E/m.
m = A x 1 uma = A x 1836 x m_e , A is the number of mass and m_e is the electron mass.
Hence:
1/β² = c²/v² = mc²/2E = A x 1836 x m_ec²/2E
We consider A = 2z, except for the proton ( where A = z ). With m_ec² = 0.511 MeV, we have:
1/β² = z x 1836 x 0.511/E = 938.20 z /E

We can expres the LET in terms of MeV x cm²/g. By dividing by the density ρ, we have for water:
1 keV/µm = 10 MeV x cm²/g.

The Bethe's formula becomes for water:
-dE/dx = 0.0170 x z²/β² {9.525 - ln[(1/β²) - 1] - β²} keV/µm.
z is the charge of the projectile, E its energy in MeV, and 1/β² = = 938.20 z /E.

6. Water LET calculator:

At the above header, a little prgram written in PHP and AJAX calculates the LET for a given projectile with charge z and initial energy E.