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 108 fast neutrons (no) 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?

The fluence Φ is given by the integral ∫∞

∞−

Φ dtdtd . The fluence rate is the number of

particles per unit area per unit time, or

dtdN

rdtd

241π

=Φ .

We can write

vdxdt = ,

where v is the speed of the train, and r, the distance from the source to a point on the train is given by

222 hxr += .

Putting it all together, we can write

( ) vdx

dtdN

hx

x

x∫+∞=

−∞= +=Φ 224

1π

.

Making use of the integral1

( )

=

+∫ hx

hhxdx arctan1

22 ,

we write ∞

∞−

=Φ

hx

dtdN

vharctan

41π

.

Evaluating the arctan function between the two endpoints gives us a factor π, so the fluence is given by

dtdN

vh41

=Φ .

1 See, for example the URL, http://integrals.wolfram.com/index.jsp

We have 1-8 sec 10=dtdN , v = 60 mi h-1, and h = 3 m, so

2-5

8

1-8

m1011.3m 0.3048 ft 5280minhrsec m 3mi 604

ftmi sec 60min 60hr 10

sec 10m 3mi 604

hr

×=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

=

⋅⋅⋅

=Φ

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 velocity2], the units for absorbed dose are [velocity2]. 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 60Co 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 × 109 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 ⋅

Ψ can be obtained from the photon fluence rate ⋅

Φ by multiplying the photon fluence rate by the energy. When we have a polyenergetic spectrum, we can sum over the energy components.

( ) ( )∑⋅⋅

Φ=Ψi

iii EEEf ,

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

( )

1-2-

2

24196

29

1-2-9

1-2-99

min m J 684min

sec 60m

cm 10eV

J106.1MeV

eV 10seccm

MeV1013.7

seccm MeV1013.7seccm MeV 107.533.15.0107.517.15.0

=

××=

×=

×⋅⋅+×⋅⋅=Ψ

−

⋅

4. (Problem 1-8 in Attix) An x-ray field at a point P contains 7.5 x 108 m-2 s-1 keV-1, uniformly distributed from 10 to 100 keV. (We might symbolize this, using ICRU#60

notation, as

=

Φ

=Φ

=ΦdadN

dtd

dEd

dtd

dEd

dEd

E

).

a. What is the photon fluence rate at P?

The photon fluence rate is found by integrating EΦ over the energy spectrum of the photons, so we write

12-10

112-8

keV 100

keV 10

secm1075.6keVsecm 107.5keV 90

−

−−

×=

×⋅=

Φ=Φ ∫ dEE

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

2-14

210

12-10

m1043.2hrsecmsec 3600hr10 6.75

hr 1secm1075.6

×=⋅⋅

××=

⋅×=

Φ=Φ−

t

c. What is the corresponding energy fluence in J m-2?

The energy fluence is determined by first integrating EEΦ over the energy spectrum of the photons to get the energy fluence rate.

( )12-12

2222112-8

keV 100

keV 10

2

keV 100

keV 10

secm keV1071.3keV 10keV 100keVsecm 107.55.0

21

−

−−

⋅

×=

−⋅×⋅=

Φ=

Φ=Ψ ∫

E

dEE

E

E

We now multiply the energy fluence rate by the time to get the energy fluence.

2-

1-31-192-16

2-16

212

12-12

m J 14.2keV eV 10eV J106.1keVm 1034.1

m keV1034.1hrsecm

sec 3600hr keV 10 3.71

hr 1secm keV1071.3

=

⋅×⋅×=

×=⋅⋅

×⋅×=

⋅×=

Ψ=Ψ

−

−

⋅

t