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

So the equivalent energy of 1 u is 1

511 keV = 931 MeV. 0:000548

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

Pr[N = k] = ake a :

k!

Its mean is given by

1 X

N = kPr[N = k] k=0

1 ake a 1 ake a X X

= k k! = (k 1)! k=0 k=1

1 a(k 1)e a 1 ake a X X

= a = a(k 1)! k!k=1 k=0

= a :

133

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134 CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE

The variance is 1 X

because 2 = E[X2] (E[X])2 : N2 = k2Pr[N = k] a2

k=0

Evaluate the summation as follows:

1 1 a(k 1)e a X X

k2Pr[N = k] = a k

(k 1)!k=0 k=1

a

(k 1) (k 1)! #= "1 + 1 a(k 1)e a

Xk=2

= a [1 + a]

= a + a2 :

So the variance of a Poisson random variable with parameter a is

N2 = a :

Solution 7.4

(a) Using (7.8), the decay constant is found as

=

0:693

=

0:693

1:4808 10 5 sec 1:

T

1=2 13 3600 secThe radioactivity A is then

A = N = 1:4808 10 5 109 = 1:4808 104 dps:

(b) Since Nt = N0e 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 exp (108 2:78 108)2 0:

p 2 2:78 108N=10 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

=

0:693

= 3:21 10 5 sec 1 :

t1=2 6 3600 sec

135

The radioactivity at t = 0 is

A0 = 1 1012 3:2 10 5 sec 1 = 3:2 107Bq = 0:86mCi :

The intensity measured is: keV

I0 = 8:91 109 :

sec m2 One hour later, the radioactivity becomes

A1 = A0e t = A0e 0:693=6 = 0:89A0 :

So the intensity measured at t = 1 hour is

I1 = 7:94 109keV

:

sec m2

Solution 7.6

(a) A0 =1 Ci= 3:7 1010Bq and At = A0e t = 1Bq. So

e t =

1

= 2:7 10 11 ;3:7 1010

which is solved as

ln 2:7 10 11 = 24:334

t = 24:334

=) t =

Since T1=2 = 0:693 = , we have = 0:693 , and t = 35:114 . It takes t = 35:114 for a radioactivesample 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 t :

The intensity at range r from this source is AtE

It=

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

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136 CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE

from the source. Therefore, the intensity on the detector face is

It(x; y) =

AtE

; jxj; jyj < D=2 :4 (R2 + x2 + y2)

(b) The average intensity is

Itav = D2 Z

D=2 Z D=2 D=2 D=2 It(x; y) dx dy

1

= D2 Z

D=2 Z D=2 4 (R2 +tx2 + y2) dx dy

D=2 D=2

1 A E

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 T1=2 :

A0 2

Taking the natural logarithm of the above equation yields

T1=2 the decay factor is

DF = e 0:693t=T1=2 :

(b) From above, we have = 1 = 0T

:1693=2 = 1:443T1=2.

= ln 2 = 0:693, and = 0:693 . So T

1=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

1:5 mCi

V =

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 disintegra-tions.

(c) Using Equation (7.11) and a = 2:6742 disintegrations (from part (b)), we have:

Prob( N > 2) = 1Prob( N = 2) Prob( N = 1) Prob( N = 0) (a)2e a (a)1e a (a)0e a

= 1

2! 1! 0!

= 1 ( a2 + a + 1)e a 2

= 1 ( 2:67422

+ 2:6742 + 1)e

2:6742 2 = 0:50003 :

Solution 7.11

Determine the decay constant of 1121Ms as follows:

1=2 = e t1=2 ;

t1=2 = 2 hours ;

= ln(1=2)

= 0:347 hr 1 :2

Determine the amount of 1121Ms left at 5 pm as follows:

t = 4 hours ; N = N0e t

= 8 g e 0:347 4

= 2 g : Subtract to determine the amount that has decayed:

8 g 2 g = 6 g :

138 CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE

Solution 7.12

(a) First determine the decay constant:

At = A0e t ;

1 mCi=ml = 3 mCi=ml e 3600 ;

= ln 1

3 36001 = 3:05 10 4s 1 :

Then find the half-life:

t1=2 =

ln 2 =

ln 2 = 2271:3 s = 0:63 h = 37:86 min :

3:05 10 4s 1

(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: 1:5 mCi

V =

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 half-life 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,

t1=2= 2:5 hours ;

=

0:693

= 7:7 10 5 s 1 ;

t1=2

N 0

= A0 = 4:4 1010= 5:19

1014 :

7:7 10 5

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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