Unformatted text preview: rons (30 × 10 Protons Ranking –3 rad)(1) = 20 × 10 –3 rem 3 –3 rad)(1) = 30 × 10 –3 rem 2 (5 × 10–3 rad)(10) = 50 × 10–3 rem 1 –3 –3 4
(5 × 10 rad)(2) = 10 × 10 rem
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Slow neutrons 7. REASONING When the cancerous growth absorbs energy from the radiation, the growth
heats up. According to the discussion in Section 12.7, the rise in temperature depends on the
heat absorbed, as well as on the mass and specific heat capacity of the growth. As we have
seen in Section 32.1, the energy absorbed from the radiation is equal to the product of the
absorbed dose and the mass of the growth. Chapter 32 Problems 1629 SOLUTION When a substance, such as the cancerous growth, has a mass m and absorbs
an amount of energy Q, the change ∆Τ in its temperature is given by ∆T = Q
cm (12.4) where c is the specific heat capacity of the substance. The energy Q absorbed is equal to the
absorbed dose of the radiation times the mass m:
Q = ( Absorbed dose ) m (32.2) Substituting Equation 32.2 into Equation 12.4 gives ∆T = ( Absorbed dose ) m Absorbed dose
Q
=
=
cm
cm
c 2.1 Gy
= 5.0 ×10−4 C°
4200 J/ ( kg ⋅ C° )
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= 8. REASONING The relation between the rad and gray units is presented in Section 32.1 as
1 rad = 0.01 gray. If, for instance, we wanted to convert an absorbed dose of 2.5 grays into
rads, we would use the conversion procedure: 1 rad = 250 rad 0.01 Gy ( 2.5 Gy ) In general, the conversion relation is 1 rad Absorbed dose ( in rad ) = Absorbed dose ( in Gy ) 0.01 Gy (1) SOLUTION
According to Equation 32.4, the relative biological effectiveness (RBE) is given by
RBE = Biologically equivalent dose
Absorbed dose (in rad) The absorbed dose (in rad) is related to the absorbed dose (in Gy) by Equation (1), so the
RBE can be expressed as 1630 IONIZING RADIATION, NUCLEAR ENERGY, AND ELEMENTARY PARTICLES RBE = Biologically equivalent dose 1 rad Absorbed dose ( in Gy ) 0.01 Gy The absorbed dose (in Gy) is equal to the energy E absorbed by the tissue divided by its
mass m (Equation 32.2), so the RBE can be written as
−2 Biologically equivalent dose
2.5 × 10 rem
=
= 0.85 6.2 × 10−3 J 1 rad E 1 rad 21 kg 0.01 Gy m 0.01 Gy ______________________________________________________________________________
RBE = 9. REASONING AND SOLUTION According to Equation 32.2, the energy absorbed is equal
to the product of the absorbed dose (AD) and the mass of the tumor:
Energy = AD × Mass = (12 Gy)(2.0 kg) = 24 J
This energy is carried by ∆N particles in time ∆t, so that 1.60 × 10−19 ∆N 6
Energy = (850 s) ( 0.40 ×10 eV ) 1 eV ∆t J Therefore, ∆N
= 4.4 × 1011 s –1
∆t
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10. REASONING According to Equation 32.2, the absorbed dose is the energy absorbed
divided by the mass of absorbing material: Absorbed dose = Energy absorbed
Mass of absorbing material The energy absorbed in this case is the sum of three terms: (1) the heat needed to melt a
mass m of ice at 0.0 °C into liquid water at 0.0 °C, which is mLf , according to the definition
of the latent heat of fusion Lf (see Section 12.8); (2) the heat needed to raise the temperature
of liquid water by an amount ∆T, which is cm∆T, where c is the specific heat capacity and
∆T is the change in temperature from 0.0 to 100.0 °C, according to Equation 12.4; (3) the
heat needed to vaporize liquid water at 100.0 °C into steam at 100.0 °C, which is mLv,
according to the definition of the latent heat of vaporization Lv (see Section 12.8). Once the
energy absorbed is determined, the absorbed dose can be determined using Equation 32.2. Chapter 32 Problems 1631 SOLUTION Using the value of c = 4186 J/(kg⋅C°) for liquid water from Table 12.2 and the
4
5
values of Lf = 33.5 × 10 J/kg and Lv = 22.6 × 10 J/kg from Table 12.3, we find that
Absorbed dose Energy mLf + cm∆T + mLv
=
=
= Lf + c∆T + Lv
in grays
mass
m
= 33.5 × 104 J/kg + 4186 J/ ( kg ⋅ C° ) (100.0 C° ) + 22.6 × 105 J/kg = 3.01 × 106 J/kg
Using the fact that 0.01 Gy = 1 rad, we find that ( ) 1 rad 8
Absorbed dose = 3.01 × 106 Gy = 3.01 × 10 rad
0.01 Gy ______________________________________________________________________________
11. SSM REASONING The number of nuclei in the beam is equal to the energy absorbed
by the tumor divided by the energy per nucleus (130 MeV). According to Equation 32.2,
the energy (in joules) absorbed by the tumor is equal to the absorbed dose (expressed in
grays) times the mass of the tumor. The absorbed dose (expressed in rads) is equal to the
biologically equivalent dose divided by the RBE of the radiation (see Equation 32.4). We
can use these concepts to determine the number of nuclei in the beam. SOLUTION The number N of nuclei in the beam is equal to the energy E absorbed by the
tumor divided by the energy per nucleus. Since the energy absorbed is equal to the
absorbed dose (in Gy) times the mass m (see Equation 32...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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