link to Problem Set 1.2 Solutions - uthgsbsmedphys.org

GS02-1093 - Introduction to Medical Physics I
Basic Interactions
Problem Set 1.2 Solutions
1. (Problem 1.7 in Attix) A point source, isotropically emitting 10 8 fast neutrons (n o )
per second, falls out of its shield onto a railroad platform, 3 meters (horizontally)
from the track. A train goes by at 60 mi h -1 . Ignoring scatter and attenuation, what is
the fluence of neutrons that would strike a passenger at the same height above the
track as the source?
d
The fluence is given by the integral dt . The fluence rate is the number of
dt
particles per unit area per unit time, or
d
1 dN
. 2
dt 4r
dt
We can write
dx
dt ,
v
where v is the speed of the train, and r, the distance from the source to a point on
the train is given by
r
Putting it all together, we can write
Making use of the integral 1
we write
2 2 2
x h .
x
1
2
4
x
dx
dN
dt
dx
.
2 xh v
1
x
,
arctan
2 2
xh h h
1 dN x
arctan
4vh
dt h
Evaluating the arctan function between the two endpoints gives us a factor , so
the fluence is given by
1 dN
.
4vh
dt
1 See, for example the URL, http://integrals.wolfram.com/index.jsp
.

We have
dN
dt
8 -1
10
sec
, v = 60 mi h -1 , and h = 3 m, so
hr
8 -1
10
sec
4
60 mi
3 m
8
10 hr 60
min 60
sec
mi
ft
4
60 mi
3 m sec
hr min
5280
ft 0.3048 m
5 -2
3.
1110
m
2. Show that absorbed dose has the same units as velocity squared. What, if anything,
does this imply about the relation of dose to velocity?
Units for absorbed dose are [energy mass -1 ]. Because the units for energy are
[mass velocity 2 ], the units for absorbed dose are [velocity 2 ]. This implied nothing
about the relation of dose to velocity because dose and velocity are two
completely different quantities; they just have the same units.
3. (Problem 1-5 in Attix) A point source of 60 Co emits equal numbers of photons of 1.17
and 1.33 MeV giving a flux density (ICRU #60 calls it “fluence rate”) of 5.7 10 9
photon cm -2 s -1 at a specified location. What is the energy flux density (ICRU # 60
now calls it “energy fluence rate”) expressed in MeV cm -2 s -1 and in J m -2 min -1 ?
The energy fluence rate
can be obtained from the photon fluence rate
by
multiplying the photon fluence rate by the energy. When we have a polyenergetic
spectrum, we can sum over the energy components.
EEE f i i i ,
i
where f(Ei) is the fraction of photons with energy Ei. In this example, the fraction
of photons with each of the two energies is 0.5. Thus
9
9
0. 51.
17
5.
710
0.
51.
33
5.
710
9
-2
7.
1310
MeV cm sec
-1
MeV cm
6
19
4
9 MeV 10 eV 1.
610
J 10 cm
7.
1310
2
2
cm sec MeV eV m
-2
-1
684 J m min
2
-2
sec
60 sec
min
-1

4. (Problem 1-8 in Attix) An x-ray field at a point P contains 7.5 x 10 8 m -2 s -1 keV -1 ,
uniformly distributed from 10 to 100 keV. (We might symbolize this, using ICRU#60
notation, as d
d d d d dN
E
dt dE ).
dE dE dt da
a. What is the photon fluence rate at P?
The photon fluence rate is found by integrating E
over the energy
spectrum of the photons, so we write
100 keV
dE
10 keV
E
8
90 keV 7.510
m
10
6.
7510
m
-2
b. What would be the photon fluence in one hour?
Because the photon fluence rate is constant the photon fluence is found by
multiplying the photon fluence rate by the time, so we write
t
sec
10
6.
7510
m
1
-2
-2
sec
sec
1
1
1hr
keV
10 hr 3600
sec
6.7510
2
m sec
hr
14 -2
2.
4310
m
c. What is the corresponding energy fluence in J m -2 ?
The energy fluence is determined by first integrating E
E over the
energy spectrum of the photons to get the energy fluence rate.
100 keV
E
EdE
10 keV
1
2
E
E
2
3.
7110
100 keV
10 keV
8
0.
5
7.510
m
12
-2
keV m
sec
-2
1
sec
keV
1
1
1
2 2 2 2
100 keV 10
keV

We now multiply the energy fluence rate by the time to get the energy
fluence.
t
3.
7110
3.7110
2.
14 J
m
12
12
1.
3410
16
1.
3410
16
-2
keV m
keVm
-2
-2
sec
1
1hr
keV hr
3600
sec
2
m sec
hr
-2
keV m
1.
610
19
J eV
-1
10
3
eV keV
-1