22.
05 Reactor Physics - Part Eight – Supplemental
                         Why Radiation Is Dangerous
1.     Background: Nuclear engineers and radiation workers all recognize that
exposure to radiation can be dangerous. But surprisingly few can explain why. A brief,
mostly qualitative, explanation is provided here.
2.      Quantification: One of the difficulties in explaining radiation, be it nuclear
power or nuclear medicine, to the public is that few people are cognizant of the units that
are employed to measure radiation. Instead, the assumption is often that there are only
two levels: 1) none which implies safety and 2) any which implies lethal. We note here
the actual units, of which there are three:
       a)      	Exposure: This is what a radiation detector measures. Radiation may
               consist of either charged (alpha, beta, protons, fission products) or neutral
               (gamma, neutron) particles. For charged particles, a detector directs the
               beam onto a gas. Each particle ionizes the gas to create an electron/ion
               pair. An electric field is imposed across the gas and that field causes the
               electron and ion to move in opposite directions to the anode and cathode
               where they are registered as a signal. Neutral particles are similarly
               detected except that they first interact with the detector wall thereby
               creating charged particles that move into the gas.
               The SI unit of exposure is the X-unit (not yet named after a person). It is
               equal to one Coulomb per kilogram (1C/kg) and is a measure of the
               amount of ionization that is created.
       b)      	 bsorbed Dose: This is the amount of energy deposited in tissue as a
               A
               result of the radiation that is incident on the tissue. Its SI unit is the Gray
               which equals one joule per kilogram. It should be noted that energy
               released is not necessarily equal to energy absorbed. When radiation
               passes through tissue, it interacts. For example, a photon may undergo a
               Compton scatter to produce an electron and a lower energy photon.
               Energy has been lost by the incident photon. But it may not be deposited
               locally unless both the scattered electron and the newly emitted low
               energy photon attenuate locally.
               To relate exposure to absorbed dose one needs to know the energy needed
               to create an ion pair. In tissue, it is 25 eV. In air, it is 34 eV. Thus, for
               air:
                       ⎛ 1C ⎞⎛⎜        1 ion   ⎞⎛ 34 eV ⎞⎛ 1.6 x 10 −19 J ⎞⎛ 1 Gy ⎞
                       ⎜        ⎟              ⎟⎜       ⎟⎜                ⎟⎜
            1 X unit = ⎜
                         kg air ⎟⎜         −19 ⎟⎝ ion ⎠⎜       1 eV       ⎟⎜ 1J / kg ⎟⎟
                       ⎝        ⎠⎝ 1.6 x10    C⎠         ⎝                ⎠⎝          ⎠
                     = 34 Gray
     c.     D
            	 ose Equivalent: This is the amount of biological damage done by the
            radiation. It is related to the absorbed dose by the relation:
                   H=QD.
            Where H is the dose equivalent, Q is the quality factor, and D is the
            absorbed dose. The quality factor allows for the LET (linear energy
            transfer) of the radiation. Photons, which interact sparsely, have a LET of
            unity. Protons, which interact densely (i.e., many nearby interactions as
            they slow down) have a LET of ten.
            The SI unit of the dose equivalent is the Sievert. It is also equal to a
            joule/kilogram because quality factor is dimensionless.
4.   Lethal Dose: The LD 50/30 is the dose that is lethal to 50% of the population in
     30 days. It is usually cited as 4.5 Sv.
5.   Energy Equivalent to a Lethal Dose: Assume (for ease of arithmetic) that a
     typical person has a mass of 100 kg. So, the energy equivalent to a lethal dose is
     450 joules. 1 J equals 1 Watt-second. So, the lethal dose is 450 Watt-seconds. In
     other words, if light from a typical reading lamp (100 W bulb) were ionizing
     radiation, you would receive a lethal dose after 4.5 seconds. The amount of
     energy involved in a lethal dose is vanishingly small. Why then is radiation so
     harmful?
6.   Interaction Mechanisms: Neutral particles generate charged particles as they
     attenuate. For example, photons produces electrons via pair production, Compton
     scattering, or the photoelectric effect while neutrons knock hydrogen nuclei
     (protons) loose. So, we need to consider how charged particles slow down. The
     energy lost per unit distance of travel is the stopping power. The figure below
     shows it as a function of particle energy:
                                   Atomic
                   Charge
                                  Collision
                   Pickup
            dE                                Radiative Loss
        −
            dx
                                                   Energy
There are three mechanisms for energy loss. At very high (relativistic) energies,
the particle radiates energy as Bremsstrahlung. At energies that are typically
encountered in most nuclear applications (1-10 MeV), the dominant mechanism is
collisions with atomic electrons. At low energies (keV and below) charge pickup
occurs – the particle becomes neutral.
How much energy can be transferred in an atomic collision? For a head-on
interaction, it is:
                    4 mM
        Q max =               E
                  (m + M )2
Where
        Q max    is the energy transferred,
        m        is the electron mass,
        M        is the mass of the incident particle, and
        E        is the energy of the incident particle.
        Consider a 1 MeV proton that strikes an electron. In this case, m<<M, and
        we obtain
                      4m
                 Q≅      E
                      M
                    ≅ 50 eV
        So, only 50 eV is lost per collision. Hence, for a single 1MeV proton to
        attenuate, 20,000 collisions are required. This is one reason why ionizing
        radiation is so dangerous. The amplification factor is enormous. There is
        another, equally important reason though. The energy lost per collision is
       on the order of 50 eV. The energy required to ionize an electron, which is
       the process whereby chemical bonds in molecule, such as DNA are
       broken, is 25 eV. Thus, the energy lost in each of the 20000 collisions that
       it takes to slow down one proton is almost exactly that needed to break a
       chemical bond.
7. 	   Effect of Other Types of Energy Sources: It remains to note why other
       energy (visible light, heat) do not have a lethal effect. The energy
       associated with visible light is typically a few eV – less than that required
       to break a chemical bond. Similarly, the energy distribution associated
       with a high temperature is the quantity kT where k is Boltzmann’s
       constant.
8.     	 ummary: To summarize, the reason that ionizing radiation is lethal if
       S
       absorbed even in seemingly small quantities is that: 1) neutral particles
       generate charged ones; 2) many thousands of atomic collisions are
       required to slow down each charged particle; and 3) the energy released
       per atomic collision is almost exactly that needed to ionize an electron in
       tissue.