Answer to Question #9658 Submitted to "Ask the Experts"

How many ion pairs/ionizations will be produced by a 1 MeV gamma ray if it passes through water/human cell? If it hits an electron and produces a photoelectric effect, how much ionization will that electron produce? What is the total mechanism of ionization?

As you likely know, gamma rays are indirectly ionizing. They possess no charge so they cannot ionize material directly. Rather they produce ionization by setting free electrons by photoelectric, Compton, or pair production events. It is these electrons that go on to produce ionization in the material.

If a 1 MeV gamma ray is incident on a single cell, it is highly likely that the gamma ray will pass through the cell with no interactions. However, if a large number of gamma rays are incident on a group of cells, as in a body tissue, many interactions may occur in the cells and the spaces between cells resulting in many secondary electrons. If we consider the case you refer to in which a photoelectric event occurs, and the resultant photoelectron enters the cell, it is highly likely that the electron will induce multiple ionization events as it traverses the cell dimensions. In order to estimate the expected number of ionization events we must know how much energy would likely be deposited by the photoelectron along its path through the cell and how much energy is required to produce an ionization event in the cellular material, which we shall assume is water-equivalent. Human cells vary widely in size, but 10 micrometers is a reasonable value to assume for this calculation.

The rate of energy loss and subsequent energy deposition by electrons is given by the collision mass stopping power, which has usual dimensions of MeV per g-cm^-2. For a 1 MeV photoelectron, the mass collision stopping power is 1.85 MeV per g-cm^-2. (You can find these values at this NIST web site.) For unit density this may also be written as energy loss per unit pathlength travelled, the linear collision stopping power, as 1.85 MeV cm^-1. If the photoelectron follows a 10 micrometer path through the cell the energy loss to the cell, E, is given by the product of the linear collision stopping power and the pathlength:

E = (1.85 Mev cm^-1)(10^-3 cm) = 1.85 × 10^-3 MeV = 1.85 × 10³ eV.

In water, approximately 50 eV of energy deposition, on average, are required to produce one ionization event (a large part of this energy goes into atomic/molecular excitation). Thus, the approximate number of ionization events, N, that we would expect from this energy deposition would be

N = (1.85 × 10³ eV)/50 eV/ionization = 37 ionizations.

Each ionization event represents the expected removal of a single electron from an atom or molecule, so the above number would represent the number of electrons, on average, expected to be set free in the cell along the path of the photoelectron.

The above assumes a photoelectric interaction, which is quite unlikely for a 1 MeV gamma ray in water. More likely such photons would interact by Compton scatter and produce a lower-energy electron. The average energy of such an electron would be about 0.44 MeV. This would not change things much, but it would increase the number of calculated ionization events from 37 to about 42 (because of a small increase in collision stopping power).

Incidentally, the energy deposition of 1.85 × 10^-3 MeV in a spherical water-equivalent cell of 10 micrometers diameter would represent a modest absorbed dose to the cell of about 0.57 mGy.

I hope this answers your question.

George Chabot, PhD