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Trang 1SOLUTIONS 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
Glenn F Knoll
Professor of Nuclear Engineering and Radiological Sciences
University of Michigan
Ann Arbor, Michigan
Trang 21-1 1-2 1-3 1-4 1-5 1-6 1-7 1-8 1-9, CHAPTER 1 Line: a,c,d,e,f,andi Continuous: b, g, and h M shell Momentum conservation: p,=p, (n represents product nucleus) p = mv = (2mE)?5 (2mE), = (2mE)n Energy conservation: E„+E„ =Q E,+(m,/m,) E, = Q E, = Q/(1+(m,/m,)) = 5.50 MeV/(1+4/206) E, = 5.395 MeV dX = hc/E dV = (4.135 x 10-15 eV-s) (3.0 x 108 m/s)/(195 x 103 eV) = 0.0636 x 10-10 m = 6.36 pm 235 —› 117%n + 118Sn Q = M(235U) c2 — M(117Sn) c2—M(118Sn) c2 M(235U) = 235.0439 M(1!7Sn) = 116.9029 M(118Sn) = 117.9016 amu Mc? = 931.5M (MeV) Q = 223.0 MeV Sp Ac (n2: Av) /„; M) = (In 2)(6.02 x 1023 mol-!) / (3.866 x 108 s)(3 g/mol) 3.60 x 10!4 Bq/g = 9.73 kCi/g E=qV=(Qe )3 MV)=6 MeV
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
Trang 32-2 2-3 energy conservation: E,+Ey.=Ey+Q or Pn2/2m, + Pye”/2Mye = Pg?/2my + Q Multiplying through by 2m, and making the approximations my, = 4m, and mg = 2m,, nˆ † PHe2/4 = pa?/2 + 2m,Q
For the neutrons emitted in the forward direction (0°), the 4He nucleus must be emitted at 180° to the incident deuteron's direction
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 E, = Pp2/2m, = (5.497)2/ 2 (1.008) = 14.99 MeV CHAPTER 2 T =1.2 x 10-7 R (m,/E) s ma=4amu E=5 MeV T = 1.073 x 10-7Rs Rg; = (5 mg/cm) / (2330 mg/cm3) = 2.1 x 103 cm = 2.1 x 105m Ts; 2.3 x 10-!2s or 2.3 ps Ry = (0.9 mg/cm2) / (0.08988 mg/cm3) = 0.1 m Ty = 1.1x10%s or 11 ns
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
AI” = 3 =0.18cm
Trang 42-5 2-6 2-7 2-8 2-9 2-10 2-11 2-12 With hv = 1 MeV, @ = 90°, and m,c? = 0.511 MeV, the formula hv’ = hv hv 1+ me (1 1—cos —cos 0 8) yields hv’ = 0.338 MeV Probability of photoelectric absorption per atom = t = const x Z*/(Ey3°) For n = 4.5, Tt Za 4.5 Si —_- 0.024 To Zq 32) Compton scattering: a, b, andd Photoelectric absorption: c Pair production: e
a) mean free path = A = 1/u = 1/[(u/p) p] = 1/[(0.06 cm2/g) (3.67 g/cm3)] = 4.54 cm
Trang 52-13 With the exposure rate const = 13.2, œ = 103 mCi, and d = 500 cm, the equation x=r< ỗ q2 yields an exposure rate of 52.8 mR/hr 2-14 E=mC,AT > AT=E/(mC,) E/m = (10 x 10-3 rad/hr) (100 erg/g-rad) (10-7 J/erg) C= 4.187 J/g-°C AT =2.39 x 10 °C/hr
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 2.22b, hg = 5 x 107! Sv-cm’ Hg = o-hg 6 « ° © = N _ (3ug)(2.3X10°neuts/s HE Nts 3600s) = 6.33x10' neuts/om? And? 4Zr(500cm) He = (6.33x10%cm™”)(5x107'Sv-cm?)(100x10°mrem/Sv) = 0.317 mrem CHAPTER 3
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
Trang 63-4 3-5 3-7 3-8 3-9 3-10 Success = rolling a six p=1/6 x=0 n=10 Using the binomial distribution, P(Q) = 0.1615 a) Success = an error p = 1/60 n= 250 mean = pn=4.17 standard deviation = [n p (1—p)]0-5 = 2.02 Using Poisson distribution, standard deviation = 2.04 b) p=1/60 n=100 x=0 Using Poisson distribution, the result is P(O) = e~100/69 = 0.189 a, b, and f net = (source + bkgd) — bkgd = 561 — 410 = 151 Onet = [ (source + bkgd) + bkgd ]05 = [561 + 410]95 = 31.2
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
T, B 73 7
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 S+B — = 84.6 | TS NT 15.46 tin 266 min ` The improvement factor is 3.03/2.66 = 1.14
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
Trang 73-11 3-12 3-13 3-14 3-15 3-16
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 a) P(5) = (5.74)5 exp(—5.74) / 5! = 0.167 b) P(>0) = 1 — P(Q) = 1 — (5.74) exp(—5.74) / 0! = 0.9968 P(>0) = 1 — P(O) = 1 — (BT)® exp(—BT) / 0! = 1 — exp(—BT) = 0.99 BT =— In 0.01 = 4.605 T = 4.605/B = 4.605/(2.87 min-!) = 1.60 min Ta = NA/TA — Ngz/Tụz = 45.1 min! Tp = Npg /Tp — Nụz/Tyz = 353.4 min'! R =rp/TA = 7.84
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“! Tg = Ny/Tr — Np/Tp Applying the error propagation formula, øs = [Nr/(Tr)2 + Ng/(Tpg)2 ]05 Øs = [rr/Tr + Npg/(Tg)2 ]?-5 = [80/T + 845/0)2 ]05 Og = 0.03 ry = 0.03 (51.833) = 1.555 Eliminating Og and solving, T = 54.1 min a) s?=0.25 [32 + 72+ 22+524+12]=22 y2 = (5—1)(22)/28 = 3.14 Using Chi-square table, p = 0.5, which indicates a good fit to the Poisson distribution b) Omean = [mean/n]Њ = [28/5]0-5 = 2.37 c) Since each of the sets is comparable to the original one, s? = 92 = 28/5 = 5.6 d) GOmean = [mean/N]® = [28/(5)(30)]95 = 0.43
For this data set, mean = 3616.1, s? = 3197, y2 = 21.22
Trang 83-17 3-18 3-19 3-20 £ =rTA —Tpg = 2162.4 - 2081.5 = 80.9 min“! e=N AT AT Np/T B Applying the error propagation formula and substituting N, for (6,4)? and Ng for (Og) : O; = [NA/ (Tạ)? + Np/ (Tp)?]85 = [rA/T At rp/T B19? 0, = [(2162.4 min-!)/(10 min) + (2081.5 min-!)/(20 min)]®5 0, = 17.9 Since the results differ by 80.9/17.9 = 4.5 o, the difference in the rates is apparently real A, =A,R where R=N,/N, (Ca/A,)? = (GAýA;)° + (Gp/R)ˆ (.02)2 = (.05⁄3.50)2 + (Gg/R)? (Gp/R)2 = 1.959 x 104 = (Øyy/N,)2 + (GØn/N,)2 = 1N, + UN, = 2 [1/000 s-)(T/2)] T = 20.4s
Trang 93-21 3-22 3-23 Using the equation N = No exp[—At], 1 N 0 N= T h N Applying the error propagation formula, and substituting No for (69)? and N, for (0;)? yields > 1 1 1 et tte" Co = + = + = 2 2 2 2 2 Not N,t Not Not Not To minimize the error, doy —2 r re” —*=0=—,(lt+e )+—> dt Not Not or 0.5Àt =1 +e-À1, Solving this equation iteratively, t = 2.218/A lọ = exp(ut) t=lIn[ly1] /H
Applying the error propagation formula and setting
Trang 103-24 3-25 a) Expected number of background counts = (100 counts/min)(30 min) = 3000 Then: oy, =V3000 = 54.8 L, =1.645V20y, =127.5 b) From Eq 3.67, Np=4.6530y_ +2.706= 257.6 From Eq 3.68, N, 257.6 œ=——= fel (0.853)(0.15)(30min)(60 / _) = 1.12/s or 1.12 Bq
Trang 114-6 a) dN dH N Ì iv 4 iv b) dN dH N 1v 4H iv 4 C) dN dH N /T\ 15V T 15V 4H 4-7 _ Resolution of the peaks requires that the energies be separated by at least 1 FWHM Let FWHM = E, — E, = 490 — 435 = 55 keV
For E,, R= FWHM/E, = 55/435 = 12.6%
For E,, R =FWHM/E, = 55/490 = 11.2%
Trang 124-10 4-11 4-12 4-13 Q = 2n(1 — d/[d2 + a2]9-5) = 2n(1 — 20/202 + 52]0-5) = 0.1876 steradians 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 a/d = tan(0.59/2) = 4.36 x 103 P = 4.76 x 10-6 N=S Cịn (2/4m S = (0.8)(20 x 103 s-!)(100 s) = 1.6 x 106 N = (1.6 x 10%)(0.12)(0.1876)/4œ = 2866 counts A = (In 2)/ty) = (In 2)/54 min = 1.28 x 10-2 min! nonparalyzable model n=m/(1-mt) Solving for t, t= m-! —n-!, Writing this result for the two observations and eliminating + yields Letting n, = nge-At and solving for no yields n= l—exp(Àt) _ 1— exp( (0.01284) (40) ) 01 1 — 1 1 m, m, 131340 93384
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 In(m,/np) _ In(m,/n,) n = 216800 0 ny
Letting n, = nge~At and solving for ng yields
Trang 134-14 4-15 4-16 4-17 3-1
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 — m„ — ,jm m; (m¡; mị) (m;; — m,) m, mM, M14 +1 yields Tt = 5.26 us
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 —
new old
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
Trang 155-6 5-8 5-10 3-11
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 I=E,,, a e/W I =(49 x 103 eV/B)(150 x 103 B sec-)(1.602 x 10-13 C/e/(27 eV/©`) = 43.6 pA I=RM At STP, M = (1.293 kg/m3)(1 liter)(10-3 m3/liter) = 1.293 x 10-3 kg/m? I = (35.8 x 10-12 C/kg-s)(1.293 x 10-3 kg/m3) =4.63 x 10-14 A V= [nge/ C] [x/d] V = [(1000 e-)(1.602 x 10-19 C/e-)/(150 x 10-12 F)] [2/5] = (0.427 nV air surroundings Ionization current vacuum surroundings | 43 g/cm? Wall thickness a) According to Table 5-2, electronic equilibrium is reached for a wall thickness of 0.43 g/cmZ
Trang 163-12 5-13 5-14 5-15 6-1 6-3 Using Table 5-2, 10 MeV gamma rays require a wall thickness of 4.9 g/cm~ t = (4.9 g/cm?)/(2.699 g/cm?) = 1.82 cm I; =RM M = (1.293 kg/m3) V [P/Pol [Ty/T] M = (1.293 kg/m3) (2.5 x 10-3 m3) [3/1] [273/373] = 7.10 x 10°3 kg Is = (100 x 10-12 C/kg-s) (7.10 x 10°3 kg) = 0.71 pA Exposure = AQ/M = AV C/M M = (1.293 kg/m?) [Vol] [P/Po] [T)/T] M = (1.293 kg/m3) [10 x 10-6 m3] [1 atm/1 atm] [273 K/293 K] = 1.205 x 105 kg Exposure = (50 x 103 V) (20 x 10-12 F) / (1.205 x 10-5 kg) Exposure = 8.30 x 108 C/kg = 0.32 mR
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
ViIn2 V
nM=T đa AV ˆK p a In(b/a)
Trang 176-5 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 ViIn2 V In(b/a) AV Kpaln(b/a) InM = yields M = 4.93 V = QIC = [Ng eMỊ/C = [(E/W}eM]/C = EeM/WC V =(5 x 106 eV) (1.602 x 10-19 C) (4.93) / [(26 eV) (500 x 10-12 F)] V =0.304 mV
The key is to make more than one measurement of output pulse amplitude for separate values of deposited energy that differ by known ratios If the amplitude remains proportional to deposited energy, the detector is operating in the proportional region Alpha particle sources often provide monoenergetic particles whose range is less than the dimensions of the chamber Because of this, the proportional counter can record each alpha particle that enters the active volume with almost 100% efficiency, and the output pulse amplitude will be proportional to the particle energy
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 Q =CV= ngeM > M= CV/ [ng e] nọ = E/W = (50 x 10 eV)/(26.2 eV) = 1908 M = (200 x 10-12 F)(10-2 V)/[(1908)(1.602 x 10-19 C)] = 6540 E=V/rin/)] -—> 1¢=V/[E In(b/a)] rc = (2000V)/[(10°V/m) In(2/0.003)] = 0.0308 cm
Venutt region’ Vtotal = [% (tc)? h] / [x b2 h] = [r¢/b]? = [0.0308/2]2 = 0.0237%
Trang 186-10
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 equation:
Vin2 In V
In(b/a) AV) K pa In(b/a)
a) Doubling the diameter of the anode wire (a) will slightly increase the first factor, but
will sharply decrease the last factor Therefore, M decreases, and V,,,,, increases
In M =
`
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
Trang 197-5
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
Trang 207-8 The critical radius for avalanche formation is obtained using E,;, = V/[r,,; In(b/a)], so
Trang 218-1 8-3 8-4 8-5 8-6 CHAPTER 8 Escint = (energy converted to light) / (total energy) =n E,/E E, = he/A = (4.135 x 10-15 eV-s) (3.00 x 108 m/s) / (447 x 10-9 m) = 2.775 eV E.cint = (20300 photons) (2.775 eV/photon) / (106 eV) = 5.63% I(t) = Ip exp[—t/T] y(t) = JT dt = [Ipt] [1 - exp[-t/t] y()/y() = 1 — exp[-t/t] = 0.99 t=T1n 100 = (230 ns) (4.605) = 1.059 ps
Maximum brightness (photons/time) occurs at time = 0, and is proportional to scintillation efficiency and inversely proportional to decay time
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
Trang 228-9 8-10 9-1 9-2 9-3 # scintillations/beta particle = (scint eff.) (Eg )/ Ephoton = (scint eff.)(Ep Y [hc/A] = (0.13) (106 eV) / [(1.24 x 10-6 eV-m)/(410 x 10-9 m)] = 43300 2 _ Th ypit — (0.15 cm), 4€ An @? View 4x (10 cm)” =5.63 x10”
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
g/cm cm
AE = 60keV
b) From 8.5, assume a light yield of 10 photons/keV Therefore, light yield = 600 photons c) From Eq 8.14, F alq_™ =1q- #2) = 0.0285 2 nạ 2 1.58 + 1m Transmission over 1 m= sxpC——) = 0.606 m Number of transmitted photons = (600)(0.0285)(0.606) =10.36 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
Trang 239-4 Emission rate of e- = l¿.„„/ gain = [2x 102 A / 105 ] (e-/ 1.602 x 10-19 C) = 12,500 e-/s 9-5 Gain = 6N = [(Vg)9-8]N = [(V/N)9-5]N = [(V/10)9-5]19 = 10-6 V6 AGain/AV = 6 x 10-6 V5 _ AV= AGain/ 6x 10-6 V° = (0.01 Gain) / 6 x 10-6 V5 = 0.01 V /6 = 0.01 (103 volts) / 6 AV = 1.67 volts 1 - 96 V@=—D—- À9(c9 An 9 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 Note that A = 1/230ns = 4.35 x 1073 ns 1/9 (ns) tmax (ns) V(tmax/(Q/C) 10 32.8 0.038 100 147 229 1000 439 645 7970 840 900 The last entry was found by trial and error to give the value 0.9 in the last column 9-7 VO)=0 in general, i(t) = CV“ + V/R
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:
RC >> T RC << T
V(t) V(t)
Trang 249-8 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
Trang 2510-2 10-3 10-4 10-5 10-6 10-7 10-8
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 v = (3.00 x 1010 cm/$)/1.85 = 1.62 x 1010 cm/s
t=d/c = (3 cm)/(1.62 x 1010 cm/s) = 1.85 x 10-10 s = 0.185 ns
This is very small compared with the 230 ns scintillation decay time Thus, the light from both events sum
In a detector of realistic size, the photopeak will be enhanced by some histories that begin with Compton scattering or pair production, but for which the full energy is eventually
absorbed Thus, the peak-to-total ratio will be larger than 1/23
R =k/[E]95
k=Re, [Ec,]2Š = (7%) [0.662 MeV]05 = 5.695
Rya = 5.695 / [Ew,]05 = 5.695 / [1.28 MeV]05 = 5.03%
Trang 2610-9
10-10
10-11
10-12
b) Calculation of backscatter again makes use of 0= 7 in nthe Compton scattering formula The results are E“(1 MeV) = 0.204 MeV
E“(2 MeV) = 0.227 MeV E“(3 MeV) = 0.235 MeV
Intrinsic peak efficiency: a andc Energy resolution: b, f, and h RC = (105 Q) (10-10 F) = 105 s
+ for Nal(TI) =230 x 102 s
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! =[{§ €, yy Qe} / (S €) & yy yo QP} - 1! = [{1/ €gy2 Qe} - 17 = [€gy2 Qe] / [1 - 22 Qe] Q= 2x (1- —==c)* 2r(1— =) = 0.663 Te d + a” J 10° + 3 Q, = O/4n = 0.0528 Given e¿ = 0.3 and y¿ = 1,
Trang 2710-13 10-14 10-15 10-16 11-1 11-2 11-3
a) The source emits positrons, which produce annihilation radiation
b) If the energy of the primary gamma rays is high, pair production within surrounding
materials can give a significant yield of annihilation radiation FWHM = R Hp = (0.085) (662 keV) = 56.3 keV
FWHM = 2.356
6 = 56.3 keV/2.35 = 23.9 keV
Low atomic number scintillator materials are used for electron spectroscopy because the probabilities of backscattering and bremsstrahlung, which are undesirable processes, increase with increasing atomic number
High atomic number scintillator materials are used for gamma ray spectroscopy because the probability of photoelectric absorption increases rapidly with atomic number
The major advantage is higher efficiency due to avoidance of sample self-aborption and attenuation by detector entrance windows
CHAPTER 11
Doping becomes significant when the dopant concentration is comparable with the intrinsic carrier concentration From Table 11-1,
For Si: 1.5 x 101° /om3 carrier concentration 4.96 x 1022 atoms/cm?
impurity level = 1.5 x 1019 / 4.96 x 1022 atoms/cm? = 3.0 x 10-13 = 0003 ppb For Ge: impurity level = 2.4 x 1013 / 4.41 x 1022 = 5.4 x 10-10 = 0.54 ppb
# carriers in silicon / # carriers in air = [E, / Wgj] /[ Ep / Wail = Wair / Ws; = (35 eV/e- ) / (3.62 eV/e- hole pair) = 9.67 Ng = E/ Wg; = (10° eV)/(3.76 eV/e- hole pair) = 26,600 e- hole pairs
Trang 2811-4 11-5 11-6 11-7 11-8 11-9 11-10 11-11 p(T) = CT” exp(-E, / 2kT) p(300 K) = C (300!-) exp(— (0.665 eV)/[2 (8.62 x 10-5 eV/K) (300 K)]) = 0.0135 C Interpolating to find E, at 77 K, E,(77 K) = 0.746 eV + (0.665 eV — 0.746 eV)(77/300) = 0.7252 p(77 K) = C (77!) exp(— (0.7252 eV)/[2 (8.62 x 10-5 eV/K) (77 K)]) = 1.27 x 1021 C Rate is reduced by p(300 K)/p(77 K) = 1.07 x 1019
The positive terminal is connected to the n side, since then one is attempting to draw electrons across the junction from the p side, where they are the minority carrier
2eV eeN 2VNe
a) d= SN b) C= 5V Cc) E ax -
(The above relationships are valid only for partially depleted configurations For fully depleted configurations, see discussion in text.)
Since the dopant concentration N appears in the denominator of Eq 11-18, the depletion depth is maximized by keeping its value small This can only be accomplished by starting with material of high purity or high resistivity
Bias voltage = 35 volts
At perpendicular incidence, 461 = K(E, — AEp), where K is the multichannel analyzer calibration in channels/unit energy
At 35°: 455 = K(Ep — AEg/ cos 35°) = K(Eg — 1.221 AE)
Subtracting second line from first: 6 =K(0.221AE)) or AE j=27.2/K Substituting this result in the first line: 461=KE )-—27.2 or K=488.2/Ey Thus: AEo = 27.2 Ep/ 488.2 = (27.2)(5.486 MeV) / 488.2 = 305 keV
The capacitance of the detector increases with surface area, adding to the noise figure at the input of the preamplifier Also, detectors with larger area also have the potential for larger leakage current at the edges of the junction
Trang 2911-12 Vuus¿= Q/C C=k (Vọp) 95 Voulse = Q(Vop)°° Mk AV ulse AV, —-045 P =0.5 (5%) = 2.5% V pulse op
11-13 Serious changes appear at 10!! a/cmZ2
Alpha flux ~ Activity/(4md2) = (10 x 106 «/s) / (4m(10 cm)2) = 7958 œ/cm2-s t= (101! œ/cm2)/ (7958 œ/cm2-s) = 1.26 x 107 s =145d 11-14 Calibration is [5.486 x 106 eV — 0]/[116-0 ch] = 47300 eV/ch Pulse height defect = true energy — apparent energy = 21.0 x 106 eV — (47300 eV/ch) (402) = 1.99 MeV 11-15 dN/dE a N\I 5 E(MeV)
Note that the alpha particles will have a dE/dx value that increases with decreasing energy, so they will lose less than half their initial energy over the first half of their range
Energy loss straggling will add width to the energy distributions in cases b and c 11-16 Will use Eq 11.6 kT T1 3 First find —— at 77K = (0.0253 V)( 555) = 6:65 x 10° V e -3 ơ II X TT 2-25) ~3.65x1072cm eE 10°V/cm = 36/0
Trang 30CHAPTER 12 12-1 R=2.35[F/NỊ05 If F=F/2, thenR,,,, = 2.35 [F/ 2N]®5 12-2 E=V/d = vụ/l = Ve/H¿ From Fig 11-2, v,,, = 107 cm/s for both electrons and holes At 77 K, py, = 4.2 x 104 cm2/V-s and pL, = 3.6 x 104 cm2/V's Using the smaller value of p, V =Ve_d/ We = Vga d/ We = (107 cm/s) (1 cm) / (3.6 x 10! cm2/V-s) = 278 V
Assume all holes and electrons must travel 10 mm to be collected Want maximum transit time < 001 T (1 cm) / (107 cm/s)(103) < T or T>104s 12-3 W,2=W,2 + W,2 + W,2 Assume charge collection is virtually complete, so W,2 = 0 W,2 = Wy? + W,2 = (2.35)? FeE + W,2 W# = (2.35)? (0.08) (2.96 eV) (140 x 103 eV) + (1.2 x 103 eV)? W, = 1.274 keV (or 0.91%) 12-4 Solving for E in the formula 2E/mụcˆ E.=E ( Ts 2m = 1.16 MeV
(with myc = 0.511 MeV) we find that E = 1.375 MeV
The location of the Compton edge (1.16 MeV) is detector independent, because it is due to interactions with free electrons
Trang 3112-6 12-7 12-8 13-1 13-2 13-3 13-4 13-5 Absolute efficiency of 7.62 x 7.62 cm Nal at 25 cm for 1.333 MeV = 1.2 x 103 (standard) Ena = (1.2 x 10-3) (25 cm/40 cm)? = 4.69 x 10-4 Ege = 0.40 Enar T=£Q¿ œ = (0.40) (4.69 x 10-4) (1.5 x 105 s-) = 28.1 s-1 W# = Wqạ? + W + W¿2 = Wq? = (2.35)2 FeE W2 = (2.35)? (0.08) (2.96 eV) (0.662 x 106 eV) W, = 930.4 eV (or 0.14%) E= 2.10 MeV —2 (0.511 MeV) = 1.078 MeV (double escape peak) CHAPTER 13
a) The fluorescent yield in Si is smaller than that in Ge
b) The incident radiation penetrates to a greater depth (photoelectric cross section in Si is small)
c) Si X-rays have low energies (1.8 keV) compared to Ge (11 keV), so their probability of escaping the detector is relatively small
tp = X/V `
v=pe=uvV/x (use H for holes, because they are less mobile than electrons)
+ {c = X2/HV = (0.4cm)2/(1.1 x 10? cm2 /V-s) (2000 V) = 7.27 ns
We should be able to resolve two peaks separated by 1 FWHM
Cu Kg = 8.028 keV Cu Ky; = 8.048 keV ZnK, =8.616keV Zn K,, = 8.639 keV
Trang 3213-6 Wp=2.35NeFE For an assumed F z 0.11 and e = 3.76 eV in silicon: W,= 2.35.|(3.76eV)(0.1 1)(59.5x10eV) = 369 eV In germanium, assume values of F = 0.08 and € = 2.96 eV: W, =2.35.|(2.96eV)(0.08)(59.5x10°eV) = 279 eV
The result expected for a silicon drift detector is identical to that shown above for a Si (Li) detector since € and F are characteristics of the detector material 13-7 The following attenuation coefficients at 662 keV were obtained from the code
Trang 3313-8 At 10 keV in Si, linear attenuation coefficient for photoelectric absorption is about 80 cm and Compton contribution is negligible (see Fig 12.22) Mean free path: aziz r =0.0125cm H 850cm Thus a thickness of 300 um absorbs nearly all incident X-rays 2 a) Readout time = (256) = 0.6555 100kHz Exposure time = (20)(0.655 s) = 13.1 s In this time, want only 5% of pixels hit, “ Tinax = (0.05)(256)(256) =250/s 13.1s b) In one interaction, deposit 4 Q= _—10 4v _ = 2.76 x10 electrons 3.62eV /e™
Since multiple hits are rare, this is minimum storage requirement c) Max leakage charge per pixel = (0 1)(2.76x10° e ) = 276 electrons
(276)(1.6 107"? C)(256)? (13.1)
Max leakage current = =2.2x10?A
CHAPTER 14
14-1 Such detectors if operated in the ionization region produce very small amplitude pulses that
are less convenient to process than those produced in the proportional region If operated in the Geiger region, all information on pulse amplitude is lost, and background or gamma ray induced events no longer can be discarded by amplitude selection
Trang 3414-5 14-6 14-7 14-8 14-9 14-10 14-11 15-1
He-3 counters can be operated at much higher pressures and are therefore preferred for those applications in which maximum detection efficiency is important
The continua caused by the wall effect are expected to be rectangular shaped only in planar geometry For tubes of small diameter, the curvature of the wall causes some particles that would otherwise deposit all their energy to deposit only part of it The effect is more noticable for the alpha particle than for the lithium particle continuum because the range of the alpha particle is larger
For interactions in the thin boron layer, there are two possibilities:
1) The alpha particle escapes the wall with no energy loss 1.47 MeV is deposited 2) The Li-7 nucleus escapes the wall with no energy loss 0.84 MeV is deposited Thus the pulse height spectrum should consist of two simple peaks of equal areas located at these two energies
Using data from Table 8.3 for excitation by fast electrons (will overestimate the actual light yield for heavy charged particles):
11,000 photons/MeV in Li(Eu), 3500 photons/MeV in Li glass
The Q-value for the °Li(n,a)°H reaction is 4.78 MeV
For Lil(Eu): N = (4.78 MeV) x (11000 photons/MeV) = 52600 photons For Li glass: N = (4.78 MeV) x (3500 photons/MeV) = 16700 photons
Once the thickness of the deposit exteeds the range of the fission fragments in the deposit material, the fragments can no longer escape from the regions farthest from the deposit surface
Number of beta particles = I/e= (10-9 C/s) / (1.602 x 10-19 C/e-) = 6.24 x 109 e-/s N(t)/No = exp[—Ogt] = exp[—(150 x 10-24 cm?)(3.0 x 1013 em-2s-1)(1.577 x 107 s)]
Trang 3515-2 15-3 15-4 15-5 a) At small diameters, the neutrons may not be moderated enough to be detected with good efficiency
b) At large diameters, the number of collisions in the moderator will tend to increase, leading to a lower value of the most probable energy when the neutron reaches the detector However, the probability that an incident neutron even reaches the detector will decrease as the moderator is made thicker
c) As the detector becomes a smaller fraction of the total volume of the system, there will be a lower probability that a typical neutron path will intersect the detector before escaping from the surface of the moderator Futhermore, a neutron may be absorbed within the moderator before it has a chance of reaching the detector, because o, is larger at lower neutron energies Counts fr Sphere Diameter `
Because virtually all reaction products are fully absorbed, the pulse height spectrum will be a simple peak centered at an energy of 4.78 MeV
Neglecting slowing down time, t = d/v = (10 cm) / (2.2 x 10° cm/s) = 45 pts This time is
long compared with electron drift times in the BF; tube (few microseconds), so the time
delay between neutron incidence and measurement of the leading edge is controlled by the neutron diffusion time
Trang 3615-6 15-7 15-8 Inserting p, = [2mE,]°> = [2 (1.009 amu) (3 MeV)]®5 = 2.46 and Q = 4.78 MeV, Pr = —3.99 Er = py2/2mr = (3.992)/ 2 (3 x 1.009) = 2.63 MeV Eq =-Er + Ea +Q =-—2.63 + 3 +4.78 = 5.15 MeV 3He+in —3H+lp
Assume helium nucleus is at rest
Energy conservation: E,+Ep=E,+Q or
pp2/2m; + pr2/2mr = pn2/2m„ + Q |
Multiplying through by m, and making the approximations m, = m, = m,/3, Pp” + Pr?/3 = p,* + 2m,Q
The maximum proton energy occurs when the proton is emitted in the forward direction (0°) and the triton is emitted at 180° to the incident neutron's direction Morhentum conservation: p, + Pr = Pp Eliminating py and solving for Pp yields 2p, +36 p, + 96m,Q P,= 3 Inserting p, = [2mE,]°° = [2 (1.009 amu) (1.5 MeV)]95 = 1.74 and Q = 0.765 MeV, Pp = 2.13 ` Ep= Pp?/2m, = (2.132/ 2 (1.009) = 2.24 MeV
The epithermal peak corresponds to the detection of incident neutrons which have been reduced to the thermal range by moderation in external materials All such neutron interactions deposit an energy equal to the Q-value of 764 keV
This type of detector makes use of elastic scattering For CHy,
NuØ
E=——— ——] 1-expl-(N,ø„ + Neøc) d ]
NuƠN + N.G,
Oy= 15 barns Oc=4.5barns d=Scm
For an ideal gas at 1 atmosphere and 300K, N = Ayn/V = AyP/RT, or N = (6.02 x 1023 g-mol-!) (1 atm) / (82.06 cm3 -atm/K/g-mol) (300 K)
= 2.45 x 101 cm Ny=4N No=N
Trang 3715-9
15-10
15-11
ER + Ene = | a (eq 1)
Ep = [4A/(1+A)?] (cos®p)* E,, , (A = 1)
Eạ = (cosÔp) Eạ o (eq 2)
Pr sinOp + ppp sin, ¿ = Ö (conservation of y-momentum)
(Pp sinÐp)2 = (bạ sinØ, ;)2
(Pp)? (1 — (cosOg)? ) = (Py, Â sinđ, ;)2
Substituting p? = 2mE, and noting that mp = m,,
Ep (1 — (cosÔg)2 ) = Ea (1 — (cosÐ, ;)2 ) (eq 3)
sing eq 1 to eliminate Eạ and eq 2 to eliminate cosÖạ, and then solving for Eạ ¿ yields
Ex ¢ = Eno (cosØ„ ;)2
E, ¢ = 1 MeV (cos 40)? = 0.587 MeV
t= x/v = x/[2E, p/m]
t = (0.03 m)/ [2 (0.587 x 106 eV) (1.602 x 10-19 J/eV)/(1.675 x 10-27 kg)]0-5 = 2.83 ns Eveiits cannot be resolved
Scattering from hydrogen is isotropic in the center-of-mass system for these energies, so we would expect each neutron group to generate an idealized rectangular response function extending from zero to the full neutron energy If each neutron group is equally intense, each rectangle will be of equal area The measured specrum will then be the superposition of these three rectangles: dN dE 75 — 150 300 Energy (keV) From the solution shown for Problem 15-9: ER † Enr= Eno
Ep = E,,.cos?@p and - Ea¿ = Enocos?20-; Thus: cos? Op + cos? @,, = 1
cos? Op = cos2 6, ¿
Trang 3815-12 15-13 15-14 15-15 15-16 16-1
Letting a = 0.005 cm, b = 2 cm, p = 0.75 atm, V = 2000 V, AV = 36.5 eV, and
K = 6.9 x 104 V/cm-atm, the equation V in2 V InM= ba) AV" Kpalinba) yields M = 5.03 V = QIC = [ng e MỊ/C = [{E/W}eM] /C = EeM / [WC] V = (106 eV) (1.602 x 10-19 C) (5.03) / [(29 eV) (60 x 10712 F)] V = 0.463 mV Ep = [4A/(1+A)?] (cosOp)? E, At 909, Ep = 0
ERlmax = [4A/+A)?] E,, = [4 (28)/(29)2] (1 MeV) = 0.133 MeV
Scattering from hydrogen at 5 MeV is isotropic in the center-of-mass system, so that ø(©) does not change with © and is equal to the constant 6, /4x Scattering from helium is not isotropic in the center-of-mass system
The radiator thickness must be kept small in order to avoid appreciable energy loss of the recoil protons before they leave the radiator
Trang 3916-2 16-3 16-4 16-5 16-6 16-7 16-8
By monitoring the reflection conditions on an oscilloscope, the termination resistance 1s
varied until no reflections occur At that point, the characteristic impedance of the cable is
equal to the termination resistance value a) rise time = 0.50 ps transit time = (20 m)/(0.659)(3.0 x 108 m/s) = 0.10 hs rise time >> transit time, so pulse is slow, and no termination is necessary b) rise time = 10 ns transit time = (10 m)/(0.84)(3.0 x 108 m/s) = 39.7 ns
rise time << transit time, so pulse is fast, and termination may be necessary
a) Use RG-58C/U with characteristic Z of 50 Q (or any other 50 Q cable) Use a shunt
resistance at receiving end of 50 Q
Trang 40The integrating factor is p = exp(/ dt / t) = exp (t/t) Multiplying the equation by p yields
dE, expt) E
Oy -exp(-/k)) exp(/t)
Integrating yields
E,, exp(t/t) = E exp(/t) -—* kot exp(—t/k) exp(t/t) + C
Att=0, E,y,=0 Therefore, C = Et/(k — 1) Dividing through by exp(t/t) yields
Ek Et
Fou = E- 7 exp(—t/k) + Cot exp(+1/T)
16-9 Want E,,,= 0.5 Ei,
Eou/Ein = |A| sin (2#ft+ 6), so |A | =0.5
[A] =[1 + Œ/Ø2]-95 =0.5
f = 0.577 f
f¡ = 1/[2wt] = 1/[2nRC] = 1/[2m(500 2)(500 x 10-12 F)] = 6.37 x 105 Hz f = (0.577)(6.37 x 105 Hz) = 3.68 x 105 Hz = 368 kHz
16-10 For this case, E,,, = E [t/t] exp(t/t) To find the amplitude of the pulse, find the
maximum Taking the first derivative and setting it equal to 0, and dividing by E, 0 = [1/t] exp(+t/t) — [t/t?] exp(—t/t) Simplifying yields t = t Therefore Eout = E /e = (1 V)/(2.718) = 0.368 V 16-11 Input to RC stage is E exp[-t/t,] The equation to solve is ful, 1p dt "a out ty exp[-/t, ]
The integrating factor is p = exp{ dt / t)) = exp (t/T,) Multiplying the equation by p, noting the left side is an exact differential, and integrating yields