Problem 7

  A beam of particles is directed at a 0.015-kg tumor. There are $1.6 \times 10^{10}$ particles per second reaching the tumor, and the energy of each particle is 4.0 MeV. The RBE for the radiation is 14. Find the biologically equivalent dose given to the tumor in 25 s.

SOLUTION: The biologically equivalent dose (BED) is defined as the product of the RBE and the absorbed dose (AD) in rads. The RBE is a number which characterizes different types of radiation in terms of their ability to damage biological material. The AD is defined as the energy absorbed divided by the mass of the absorbing material. We are given the RBE. We need to find the AD. We know the mass of the absorbing material, so we need to find the energy absorbed. We know the energy per particle, the number of particles per second, and the total time; the energy absorbed is these three things multiplied and converted to SI units (grays)

$$E_{abs} = (4 eV)(1.6 \times 10^{10} s^{-1})(25 s) \left ( \frac{1.60 \times 10^{-19} J}{1 eV} \right )$$

$$AD = \frac{E_{abs}}{0.015 kg} = 17 grays = 1.7 \times 10^3 rads$$

where 1 gray = 100 rads. The BED is then

$$BED = (1.7 \times 10^3 rad)(14) = 2.4 \times 10^4 rem .$$