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SOLUTIONS TO PROBLEMS

from

Radiation Detection and Measurement,

Third Edition, by Glenn F Knoll

Published in 2000 by John Wiley and Sons, New York

These solutions have been prepared with the assistance of a number of past and present students from the Department of Nuclear Engineering and Radiological Sciences

at the University of Michigan, with major contributions from Paul Nowak, Richard

Kruger, Jim Fox, and Shana Browde The Mathematica solutions in Chapter 17 were provided by Prof David Wehe

Heavy water contains deuterium For 27H(y,n), Q = —2.224 MeV

Therefore, the minimum gamma ray energy is 2.224 MeV

?2H+3H —›4He+ln

Assume tritium nucleus is at rest

Momentum conservation: p, + Pye = Pg

Eliminating py, and solving yields: p, = (pg + [6py2 + 40m,Q]°> )/5

Inserting pg = [2mgEg]®> = [2 (2)(1.008 amu) (0.150 MeV)]°-5 = 0.778

and Q = 17.6 MeV leads to p, = 5.497

R = 210 ym for 5 MeV protons in Si Therefore, the range remaining

after 100 hm is (210 — 100) or 110 pm Corresponding energy remaining = 3.2 MeV

The expected energy loss is given by AE = (-dE/dx)-t

From Fig 2-10, (-dE/dx)/p for a 1 MeV a particle is approximately 380 MeV-cm2/g Pau = 19.32 g/em3 =t=5x104+cm

Using these values, AE = 3.67 MeV Therefore, the o particle loses all its energy

According to the figure, (PR)s; = 0.5 g/em2 Using the scaling law,

— PsRs; Am — (05 g/em2 /2698

Paral Ag (2.699g/cm)./28.09

a) mean free path = A = 1/u = 1/[(u/p) p] = 1/[(0.06 cm2/g) (3.67 g/cm3)] = 4.54 cm

b) prob of photoelectric absorption = 1 — exp[—T t]

2-13 With the exposure rate const = 13.2, œ = 103 mCi, and d = 500 cm,

2-15 From Chapter 1 (p 20), we know that 2.3 x 10° neutrons/s are produced per pg of

*°2Cf and that average neutron energy is between 0.5 and 1 MeV So from fig

3-1 | To first approximation, the sample variance is independent of the size of the sample

Therefore, there should be no significant change in its value

3-2 Using the binomial distribution, with success defined as "heads",

count rate = [(source + bkgd) — bkgd]/ time = (846 — 73)/(10 min) = 77.3 min—!

= [ (source + bkgd) + bkgd ]05 / time = [ 846 + 73 ]05 / (10 min) = 3.03 min-l

Ốcount Tate

Using the data from prob 3-8, optimal allocation is given by

TB S+B _ 84.6 _ 340

Since Top + Tp = 20 min, Tsip = 15.45 min, and Tp = 4.55 min

With this time allocation, the standard deviation in the net source counting rate is

As a fraction, o = [1/N]®-5 = [1/ST]°5, where N is the number of counts, S is the count

rate, and T is the time interval Since the count rate is constant,

[o? T], = [0% T]o, or

T\¡ = Tạ [ØgG]2 = (10 min) [2.8/1.0]? = 78.4 min

The additional time = 78.4 —10 = 68.4 min

a) If S >> B, the figure of merit 1/T = €2S Therefore, double the source signal (S)

b) If S <<B, 1/T = e2S2/4B Therefore, double S

mean = (2.87 min-!) (2 min) = 5.74

This value is small, so use Poisson statistics

To calculate og, use R = [Np/Tg — Nụ /Tp]/[NA/TẠ — Nụ„/Tp,]

Application of the error propagation formula yields Og = 0.634

Tp = 845/(30 min) = 28.17 min-} Ty = 80 — 28.17 = 51.83 min“!

For this data set, mean = 3616.1, s? = 3197, y2 = 21.22

Using Chi-square table, p ~ 0.5 They are consistent with the Poisson distribution

Tị = Tạ —Tpẹ = (1683/10 min) — 50 min“! = 118.3 min“!

T> = Tr —T = (914/10 min) —50 min“! = 41.4 min“!

o2(r,) = 02(1,) + o2(r,) = [(1683)1/2/10)]2+0 = 16.83

O*(ry) = O2(rp) + O2(r,) = [(914)1/2/10)]2 +0 = 9.14

R= 1o/r, = 118.3/41.4 = 2.857

(Gp/R)ˆ = [Ø(s)/rạ] + [G(rq)/r]2 = 9.14/(41.4)2 + 16.83/(118.3)2 = 6.535 x 103 Half-life = T =0.693t/n R = 0.693(24h)/In 2.857 = 15.84h

Applying the error propagation formula and setting

(0;)2 =I =Ip exp( ut) = (Ip and t have 0 variance),

Np=4.6530y_ +2.706= 257.6 From Eq 3.68,

b) Assuming a Poisson random process, t=1/r= 1/(30min)"' = 30 min

For E,, R= FWHM/E, = 55/435 = 12.6%

For E,, R =FWHM/E, = 55/490 = 11.2%

Taking the smaller value, R must be smaller than 11.2%

4-8 R=2.35[F/N]5

N =5.5225 F/R2 = (5.5225)(0.1)/(0.005)2 = 22 090

P=Q/4x (assuming all space is randomly sampled)

= 2n(1 — d/[d2 + a2]9-5)/4x = (1 — d/[d2 + a2]95)/2 = (1 — 1/[1 + (a/d)2]95)/2

paralyzable model m =n exp(-nt)

Solving for —t, —t = (In[m/n])/n Writing this result for the two

observations and eliminating Tt yields

Letting n, = nge~At and solving for ng yields

At-e™™* In(m,) + In(m,) n,, = ex

The dead time loss in A = n — mag, and dead time loss in B = n — mạ

We require n — mg = 2(n—m,) Solving for n, n = 2m,— mg

ma, =n/[1+nt,] and mpg =n/[1+ntTp] Substituting these expressions

into the expression for n and again solving for n yields

1 2

n= — —

UA th

Letting ta = 30 x 1Š s and tp = 100 x 106 s yields n = 13 333 s-1

With mị = m; = 10 000 s-! and m,, = 19 000 s-!, the equation

Paralyzable model: m =n exp(-nt), or 0 = n exp(—nt) —m

Using the Newton-Raphson method, the latter representation yields

"¡am exp[n „ qu

=n —

Letting m= 10’ s-! andt = 1.5 x 10-*s, an initial guess of 0 yields n = 119,700 s-! and

an initial guess of 107 yields n = 1,996,000 s-1

max = L/Et e]

t = 1/(m,,,x €] = 1/[(S0 000 s-!) (2.718)] = 7.36 ps

CHAPTER 5

# electrons = E/W = (5.5 x 106 eV)/(42.7 eV/ion pair) = 1.29 x 105

Q = (# charge carriers)(charge per carrier)

Q = (1.29 x 10°)(1.602 x 10-19 C/e-) = 2.06 x 10°14 C

I = (# a's per second)(charge per a) = (300 s-!)(2.06 x 10-14 C) = 6.19 pA

Plot for Problem 5-2:

Using the Si curve in Fig 2-14, , (pR)s; = 0.25 g/cm¿

Re RyPsinf a Anz (025gm)/2897 - 211ơn = : -

" PainfAs (1.204 10° g/em*)/28.09

Secondary electrons created by the 5 MeV gamma rays must not reach either electrode For minimum electrode spacing, assume the electrons are created exactly in between the parallel electrodes For 5 MeV gamma rays, typical secondary electron ranges are on the order of several meters Consequently, the electrode spacing would have to be larger than several meters for compensation to be preserved

b) In a vacuum, no secondary electrons enter the chamber, so there is inadequate

compensation using thin walls This reduces the measured ion current compared with air surroundings

First portion: V = [nge/dC] (v†+v-)t —> Ví =[nge/dC] (v† + v-)

Second portion: V = [nge/dC] (v†t+x) —> Ví =[nge/dC] (v?)

R = (slope of first portion)/(slope of second portion) = 1 + v-/V†

v=u-E and v†=k+E

R =1 +ưíur > 1 + 1000 = 1000

CHAPTER 6

n = E/W = (106 eV)/(26.2 eV/ion pair) = 3.82 x 104 ion pairs

Ơn = [F n]?* = [(0.17) (3.82 x 10%]0-5 = 80.6 (or 0.21% of n)

Air is not used in proportional counters because it has a substantial electron attachment

coefficient, so that gas multiplication cannot occur under normal conditions Air can be

used in ion chambers because either free electrons or negative ions may then be collected

a) Letting M = 1000, a = 0.003 cm, b = 1 cm, p = 1 atm, AV = 23.6 eV,

and K = 4.8 x 104 V/cm-atm, the equation

nM=T đa AV ˆK p a In(b/a)

yields V = 1793 volts

6-5

6-6

6-7

6-8

_ b) Lettinga =0.006 and V = 1793, the equation yields M = 7.52, so that the

multiplication decreases by a factor of 1000/7.52 = 133

c) Letting a = 0.003, b= 2, and V = 1793, the equation yields M = 192, so that the

multiplication decreases by a factor of 1000/192 = 5.2

Letting a = 0.005 cm, b = 5 cm, p = 1 atm, V = 2000 V, AV = 23.6 eV, and K = 4.8 x 104 V/cm:atm, the equation

In contrast, the range of beta particles exceeds the chamber dimensions The number of ion pairs, and hence the pulse height, is only proportional the the small fraction of particle energy lost in the chamber This fraction decreases as the beta particle energy increases, because the rate of energy loss, dE/dX, decreases as the beta particle energy increases Therefore, the higher the energy of the beta particle, the smaller the deposited energy, and

the smaller the pulse height

Venutt region’ Vtotal = [% (tc)? h] / [x b2 h] = [r¢/b]? = [0.0308/2]2 = 0.0237%

The field tube prevents gas multiplication from occurring near the ends of the counter, where the electric field is distorted

6-10

6-11

7-1

7-3

The discontinuities are absorption edges corresponding to the binding energies of electrons

in K, L, and M shells in the absorber atoms If the gamma energy is above an absorption edge, photoelectric absorption can occur If the gamma energy is slightly below the edge, this process is not energetically possible, and the total interaction probability drops

abruptly

The fluctuation in the size of individual avalanches is avoided, and the statistical limit on energy resolution is lower

CHAPTER 7

The quench gas has the lower ionization potential in order to favor charge transfer

collisions in which the positive ions of the primary fill gas are neutralized, and positive ions of the quench gas are collected

The starting voltage reflects the point at which the avalanche size is sufficient to assure spread of the Geiger discharge The avalanche size is proportional to the multiplication factor M, so the starting voltage should vary inversely with M From the Diethorn

b) Doubling the fill gas pressure (p) will decrease the last factor Therefore, M decreases,

and V tart increases

c) Consider the change from P-5 to P-10 gas This results in increased values for both K and AV This results in a lower value for M and, therefore, a higher value of Vua:

Increasing the voltage results in a higher initial electric field, so that a greater buildup of space charge is required to reduce the field below its critical value Consequently, the pulse height increases

To reach the Geiger region, (ng)’p 2 1

(no)’ =3M

M 2 1/3p = 1/3 x 10°) = 3.33 x 10

Using the equation and data in prob 6-3 a), setting M = 3.33 x 104 and iterating, the result is V= 2166 volts

7-5

T-6

T-1

a) Proportional: The pulse height varies as the avalanche amplitude which, in turn,

depends on voltage in an approximately exponential manner

Geiger: The pulse amplitude corresponds to the number of ion pairs at the point at which the accumulated positive space charge is sufficient to reduce the electric field below its critical value This number will increase in approximate proportion to the original electric field or linearly with the applied voltage

b) Proportional; The quench gas must absorb UV photons

Geiger; The quench gas must pick up positive charges from the original positive ions

through charge transfer collisions

c) Proportional: Because heavy charged particles tend to deposit all of their energy, and electrons only part of theirs, the two radiations can be separated by their different pulse heights

Geiger: No differentiation can be achieved, because pulse height is independent of particle type and energy

d) Proportional: The maximum counting rate is often set by pulse pile-up The minimum pulse shaping time (that will minimize pile-up) is limited by the finite rise time of the pulses

Geiger; The maximum counting rate is limited by the long dead time of the tube itself e) Proportional: Gamma rays produce very small amplitude pulses and are often below the discrimination level

Geiger: Counting efficiency is a few percent due primarily to the liberation of

secondary electrons from the detector walls

a) The response to gamma rays is due to interactions in the solid wall of the detector

These interactions result in the liberation of secondary electrons that ionize the gas The free electrons so formed subsequently drift to the anode wire and create an avalanche The

efficiency for counting gamma rays depends on a) the probability that the incident gamma ray interacts in the wall to produce a secondary electron, and b) the probability that the secondary electron reaches the fill gas before the end of its track Process a) generally increases with wall thickness Process b) is initially a weak function of wall thickness, but decreases when the distance of the average point of interaction from the inner wall exceeds the secondary electron range With large wall thickness, attenuation of the incident gamma rays ultimately decreases the efficiency

b) t,, is typically 1-2 mm, and is approximately equal to the maximum secondary electron

range

If another event occurs during a dead time, it can trigger another Geiger discharge, but of smaller then normal amplitude Thus it normally will not be counted, but will extend the time needed to clear the tube of excess positive charges The tube will then behave more closely as a paralyzable detector

7-8 The critical radius for avalanche formation is obtained using E,;, = V/[r,,; In(b/a)], so

Tort = W/LE git In(b/a)] = 1500 V/[(2 x 106 V/m)(In(2/0.005)] = 1.25 x 10-2 cm

at nn m=ne `” =—

2

Solving: m _n2_ 9633 _1980/s

tT 35005 7-10 a) Setr-t=0.05 since r-t<< 1

Ratio = [€,cin¢(Nal/ €,cin Canth)][t(anth)/ t (NalI)] = [230/100][30ns/230ns] = 0.30

Thus, even though the light yield from anthracene is less, its maximum brightness is greater than that for NalI(TI)

Organics generally have faster response (although several fast inorganics are in use) Inorganics have higher light output, are more linear, and have higher detection efficiencies for high energy gamma rays

The cost depends on the application, but generally favors plastic or liquid organics for large volume detectors

Activators are used in inorganic scintillators in order to increase the probability of visible photon emission Transitions across the band gap of the parent crystal tend to yield

photons with energies higher than for visible light, and these photons can be reabsorbed

Activators introduce intermediate energy levels and photons that can be transmitted

efficiently through the crystal They are not required in organics because the scintillation process arises from transitions within the energy level structure of individual molecules Nal(T]) has the highest scintillation yield when measured using standard PM tubes, but CsI(TI) has a higher yield when measured using photodiodes with broad spectral response The scintillation process in organics arises from transitions within the energy level

structure of a single molecule, and can be observed from a given molecular species

independent of its physical state Inorganic scintillators, however, require a regular

crystalline lattice as a basis for the scintillation process

8, = sin“! [Ngiy/Nplastic] = sin"! [1.0/1.58] = 39.279

photons received per flash = (# scintillations/beta particle)(Q, ,,) = 2.4

- Flash is not visible

a) Since a 1 MeV electron is a “minimum ionizing particle” (see Fig 2.1), its energy loss is about 2 MeV per g/cm’ in light materials (see p 32)

Energy loss AF= He -1.03—5— - 0.3mm

AE = 60keV b) From 8.5, assume a light yield of 10 photons/keV

Therefore, light yield = 600 photons

CHAPTER 9

Ep =he/A = (1.24 x 10-6 eV-m) / (1.5 eV) = 827 nm

Vavg = 0.5 [2Er/m]1⁄2 = 0.5 [2(150 eV)(1.602 x 10-19 J/eV) / 9.11 x 10-31 kg]1⁄2

= 3.63 x 105 m/s

t = X/Vayg = 12x 10-3 m/ 3.63 x 106 m/s = 3.30 ns

e~ gain = 106 = 8 — 8 = 10 — E(e-) = 200 eV > V = 200 volts per stage

Overall voltage = (200V/stage)(6 stages) = 1200 V

9-4 Emission rate of e- = l¿.„„/ gain = [2x 102 A / 105 ] (e-/ 1.602 x 10-19 C)

To find the time at which the pulse reaches its maximum, differentiate V(t) with respect to

time, and set equal to zero Solving for this time, t,,,, = [1/(@-A)]In(6/A) Inserting this value for t gives the maximum amplitude for a given time constant (1/8)

The “amplitude reached by the pulse using an infinite time constant is just Q/C

for0<t<T,I{=I — I=CVf+V/R > V(Œ€)=RI(1—-e-RQ

for T<t,iQ=0 —> O=CV’+V/R —> V() = RI (1 — e1/RQ) e-RC

Sketches:

9-8 RC=(100)10'°F)=107s

+ for Nal(T)) = 230 x 107 s

Since RC >> t, the maximum pulse height V = Q/C

n = E-€scin/hV = (1.2x10° eV)(0.12)/(3 eV/photon) = 4.8x10* photons

Q = n(light coll eff.)(photo cath eff.)(1" dynode eff.)(gain)qe

= (4.8x10*)(0.7)(0.2)(0.8)(100,000)(1.62x10"? C) = 8.71x10"' C

max pulse height = Q/C = (8.71x10"' C)/(10”° F) = 0.871 V

9-9 | Microchannel plate PM tubes show excellent timing properties The total electron transit

time through a channel is a few ns, compared with 20 — 30 ns for conventional multiplier structures The spread in transit time, which determines timing performance, is about

100 ps, a factor of two or three better than conventional PM tubes

9-10 A=hc/E = (1.24 x 105 eV-m)/(1.11 eV) = 1.12 im

Energy deposition rate = (5x10° eV)(10°) = 5x10! eV/s

Using the formula for Compton scattering, with E = 2 MeV and 8 = 309, E“= 1.312 MeV

Using the formula again, with E = 1.312 MeV and 8 = 609, E’*= 0.574 MeV

The energy deposited = 2 — 0.574 = 1.426 MeV

This result is independent of the sequence of scattering

For any event, maximum energy deposition occurs for @ = 180°, so that

E E=—————~ 1 + (2E/m,c’)

A | MeV photon will yield a 0.204 MeV photon, which in turn will yield a 0.113 MeV

photon The energy deposited = 1—0.113 = 0.887 MeV

For Nal, the index of refraction n = c/v = 1.85

k=Re, [Ec,]2Š = (7%) [0.662 MeV]05 = 5.695

Rya = 5.695 / [Ew,]05 = 5.695 / [1.28 MeV]05 = 5.03%

€;, = [# of pulses recorded] / [# of pulses incident] = 1 — expl-(H/p)pt]

Since RC >> t, the maximum pulse height V = Q/C

n= E-€,.;,/hv = (10° eV)(.12)/(3 eV/photon) = 40 000 photons

Q =n (light coll eff.) (photo cath eff.) (1st dynode eff.) (gain) q,

Q = (4.0 x 104) (0.5) (0.2) (0.8) (2.519) (1.602 x 10-19 C) = 4.89 x 10-12 C

max pulse height = Q/C = (4.89 x 10-!2 C)/(10-!0 F) = 48.9 mV

a) [counts sum]/[counts y, peak] = N,2/[N, — Nya] =[ N,/Nj2—- 11!

Q, = O/4n = 0.0528 Given e¿ = 0.3 and y¿ = 1,

[counts sum]/[counts y, peak] = 0.0161