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7 The Physics of Nuclear Medicine
FUNDAMENTALS OF ATOMS Solution 7.1 The mass of an electron me is 0.000548 u. So 1u =
1
me. The equivalent energy of an electron is 511 keV.
0:000548
1
So the equivalent energy of 1 u is
0:000548
511 keV = 931 MeV.
Solution 7.2 The mass defect of a deuteron is 1:007276+1:008665 2:01355 = 0:002391u. Its binding energy is 0:002391 u 931 MeV/u= 2:228 MeV.
RADIOACTIVE DECAY AND ITS STATISTICS Solution 7.3 The PMF of a Poisson distribution with parameter a is given by
ake a :
Pr[N = k] =
k!
Its mean is given by X
1
N =
kPr[N = k] k=0
=
X
ake a
1
k=0 1
= a = a:
=
k k! X
k=1
a(k
(k
ake a
1 X
k=1
(k 1
1)e a
1)!
133
1)! X
=a
k=0
ake a
k!
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CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE
The variance is
1 X
2 N =
because 2 = E[X2] (E[X])2 :
k2Pr[N = k] a2
k=0
Evaluate the summation as follows: 1 X
X
a(k 1)e a
1
2
k Pr[N = k] = a k (k k=1 a "1 + =
k=0
1)!
1)! #
(k 1) (k
1
a(k
1)e a
X
k=2
=
a [1 + a]
= a + a2 : So the variance of a Poisson random variable with parameter a is 2 N =a:
Solution 7.4 (a) Using (7.8), the decay constant is found as 0:693
T
= The radioactivity A is then
1=2
0:693 5 1 = 13 3600 sec 1:4808 10 sec :
10 5
A = N = 1:4808 (b) Since Nt = N0e
109 = 1:4808 104 dps:
t, then
N24 h = 109 exp( 1:4808
10 5
24
3600)
2:78
108 atoms:
(c) The number of radioactive atoms left follows a Poisson distribution with a mean as computed in (b). For large mean value, the Poisson distribution can be well approximated by a Gaussian distribution with the same mean and variance. Thus, P
8
1
=
p
N=10
exp
2 2:78 108
(108 2:78 108)2
0:
2 2:78 108
Solution 7.5 At t = 0, the number of technetium-99m atoms is 1 1012. Since the half-life of technetium-99m is 6 hours (Table 7.1), the decay constant is 0:693 =
t
1=2
0:693 5 1 =6 3600 sec = 3:21 10 sec :
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135
The radioactivity at t = 0 is A0 = 1
1012 3:2
10 5 sec 1 = 3:2
The intensity measured is:
107Bq = 0:86mCi :
keV
I0 = 8:91 109 One hour later, the radioactivity becomes
sec m2 :
A1 = A0e t = A0e 0:693=6 = 0:89A0 : So the intensity measured at t = 1 hour is keV I1 = 7:94 10
9
sec m2 :
Solution 7.6 (a) A0 =1 Ci= 3:7 1010Bq and At = A0e t = 1Bq. So 1 10 11 ;
e t = 3:7 1010 = 2:7 which is solved as ln 2:7 t = 24:334
10
=
11
24:334
=) t = Since T1=2 = 0:693 = , we have = 0:693 , and t = 35:114 . It takes t = 35:114 for a radioactive sample with activity 1 Ci to decay to activity 1 Bq if the half-life is . (b) The radioactive tracers used in nuclear medicine should have a half-life on the order of minutes to hours, about the time it takes to perform study. If longer, activity remains in patient. If shorter, activity disappears before scan is completed.
Solution 7.7 (a) The radioactive source decays according to At = A0e The intensity at range r from this source is =
t
:
I AtE t
; 4 r2 where the time-dependency is made explicit using a subscript t and E is the gamma-ray energy. A point (x; y) on the detector is at a distance p
r=
R2 + x2 + y2
136
CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE
from the source. Therefore, the intensity on the detector face is AtE 2
4 (R + x2 + y2)
It(x; y) = (b) The average intensity is Itav = =
D2
Z D=2
1
D2
D=2
Z
1
D=2 D=2
Z
; jxj; jyj < D=2 :
D=2
D=2 It(x; y) dx dy
Z
D=2 D=2
4 (R2 +tx2 + y2)
dx dy
AE
AtE 4 R2 ; where the last approximation holds if R D.
Solution 7.8 (a) DF is defined as DF = e t. And decay constant is given by
A1=2 1 = =e A0
T1=2
:
2
=
Taking the natural logarithm of the above equation yields T1=2 the decay factor is DF = e
0:693t=T
1=2
ln 2 =
0:693, and = 0:693 . So T1=2
:
(b) From above, we have = 1 = 0T:1693=2 = 1:443T1=2.
Solution 7.9 (a) The half-life of 99mTc is 6 hours. It is 8 hours from 8 a.m. to 4 p.m. Therefore, using the relation between the decay constant and the half-life, = 0:693
T
1=2
we can write
A4p:m: = A8a:m:e t = 2e 0:693 8=6
0:7939 mCi/ml :
(b) To get 1.5 mCi radioactivity, we need a volume of
V=
1:5 mCi 0:7939 mCi/ml
1:89 ml :
137
Solution 7.10 (a) First we find the decay constant as follows Nt = N0e t 9:9212 106 = 108e 864000 = ln((9:9212 106)=108)=864000 = 2:6742 10 6 sec 1 : Using the relationship between the half-life and the decay constant, we find t1=2 = =
ln(2)= 259198 259200 sec (or 3 days)
(b) For t << t1=2, the average number of disintegrations = Poisson rate t = N0 t = 2:6742 disintegrations. (c) Using Equation (7.11) and a = 2:6742 disintegrations (from part (b)), we have: Prob( N > 2) = 1 Prob( N = 2) Prob( N = 1) Prob( N = 0) (a)2e a
= 1 =
1(
a2
= 1( =
(a)1e a
2!
(a)0e a
1!
+ a + 1)e a 2
2:67422
2:6742 2
+ 2:6742 + 1)e
0:50003 :
Solution 7.11 Determine the decay constant of 1121Ms as follows:
1=2 = e t ; 1=2
t1=2 = 2 hours ; =
0!
ln(1=2)
2 21 Determine the amount of 11 Ms left at 5 pm as follows:
= 0:347 hr 1 :
t = 4 hours ; N = N0e t = 8 g e 0:347 = 2g:
4
Subtract to determine the amount that has decayed: 8g
2g=6g:
138
CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE
Solution 7.12 (a) First determine the decay constant: t
At = A0e
;
1 mCi=ml = 3 mCi=ml e =
ln
1
3600 ;
3 3600
1
= 3:05 10 4s 1 :
Then find the half-life:
t
1=2
=
ln 2
=
ln 2 3:05 10 4s 1
= 2271:3 s = 0:63 h = 37:86 min :
(b) Compute the radioactivity: At = 3 mCi=ml e
4 3600
= 3 mCi=ml e 3:05 10 4s 1 4 3600s = 0:6371 mCi=ml : (c) Calculate the volume:
V=
1:5 mCi
0:6371 mCi=ml
= 2:3544 ml :
RADIOTRACERS Solution 7.13 (a) Explanation for each: (i) E = 30 Kev, t1=2 = 7 hours: This is a bad choice for medical imaging because the energy of the gamma rays are low and the body will absorb most of the emitted gamma rays. (ii) E = 150 Kev, t1=2 = 5 hours: This is a good choice for medical imaging purposes because its half-life is long enough to enable imaging and short enough to weaken strongly before the patient leaves the hospital. The gamma ray energy is high so that it is somewhat transparent in the body but still detectable by conventional detectors. (iii) E = 200 Kev, t1=2 = 10 days: The energy would be a pretty good choice for this one. The halflife would be good for biological processes that take a week or so for the radiotracer to reach its destination. It is too long, however, for most processes. (b) Activity follows the radioactive decay law. If activity reduces to 1/4 after 5 hours then 5 hours is twice the
139
half life. Accordingly,
t
1=2 =
2:5 hours ; 0:693
t
= 7:7 10 5 s 1 ; 10 = A0 = 4:4 10 = 5:19 =
N 0
1=2
7:7
10 5
1014 :
Solution 7.14 A radiotracer is chosen first for its properties of biodistribution, and then its physical imaging properties. The two radiotracers are not equivalent if they distribute in the body in different ways and most likely they cannot be interchanged.
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