1.2-ProblemSetSolutions

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

dΦ dt . The fluence rate is the number of dt −∞

∫

particles per unit area per unit time, or dΦ 1 dN . = dt 4πr 2 dt

We can write dt =

dx , 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 2 = x2 + h2 .

Putting it all together, we can write x = +∞

∫

Φ=

x = −∞

(

1

4π x + h 2

2

)

dN dx . dt v

Making use of the integral 1

∫ (x

2

dx 1  x = arctan  , 2 h +h h

)

we write ∞

1 dN  x Φ= arctan  . 4πvh dt  h  −∞

Evaluating the arctan function between the two endpoints gives us a factor π, so the fluence is given by Φ= 1

1 dN . 4vh dt

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

We have

dN = 108 sec -1 , v = 60 mi h-1, and h = 3 m, so dt

hr ⋅108 sec -1 4 ⋅ 60 mi ⋅ 3 m 108 hr ⋅ 60 min ⋅ 60 sec ⋅ mi ⋅ ft = 4 ⋅ 60 mi ⋅ 3 m ⋅ sec ⋅ hr ⋅ min ⋅ 5280 ft ⋅ 0.3048 m = 3.11×105 m -2

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. ⋅

⋅

Ψ = ∑ f (Ei )Ei Φ (Ei ) , 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 ⋅

(

)

Ψ = 0.5 ⋅1.17 ⋅ 5.7 ×109 + 0.5 ⋅1.33 ⋅ 5.7 ×109 MeV cm -2sec -1 = 7.13 ×10 MeV cm sec 9

-2

-1

MeV 106 eV 1.6 × 10 −19 J 10 4 cm 2 60 sec cm 2 sec MeV eV m2 min -2 -1 = 684 J m min = 7.13 ×109

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   = dΦ = d  dΦ  = d  d  dN   ). notation, as Φ E dE dE  dt  dE  dt  da   a. What is the photon fluence rate at P?  E over the energy The photon fluence rate is found by integrating Φ spectrum of the photons, so we write 100 keV

∫ Φ

= Φ

E

dE

10 keV

= 90 keV ⋅ 7.5 ×108 m -2 sec −1 keV −1 = 6.75 ×1010 m -2 sec −1

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 Φ=Φ = 6.75 ×1010 m -2 sec −1 ⋅1 hr hr × 3600 sec = 6.75 ×1010 m 2 ⋅ sec⋅ hr = 2.43 ×1014 m -2

c. What is the corresponding energy fluence in J m-2?  E over the The energy fluence is determined by first integrating EΦ energy spectrum of the photons to get the energy fluence rate. 100 keV

⋅

Ψ=

∫ EΦ

E

dE

10 keV

100 keV

=

1  2 ΦEE 2 10 keV

(

= 0.5 ⋅ 7.5 × 108 m -2 sec −1 keV −1 ⋅ 100 2 keV 2 − 10 2 keV 2 = 3.71×10 keV m sec 12

-2

−1

)

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

Ψ = Ψt = 3.71×1012 keV m -2 sec −1 ⋅1 hr keV ⋅ hr × 3600 sec = 3.71 ×1012 m 2 ⋅ sec⋅ hr = 1.34 ×1016 keV m -2 = 1.34 ×1016 keVm -2 ⋅1.6 ×10 −19 J eV -1 ⋅103 eV keV -1 = 2.14 J m -2