moe section 2

8/13/2019 Moe Section 2

1/32

SECTION 2 - RADIOACTIVITY AND ITS PROPERTIESA. Earlv Historv

Henri Becquerel is credited with the discovery of natural radioac-tivity. While working with uranium salts and photographic plates in 1896,he found that the uranium emitted a penetrating radiation which seemedsimilar to that which Roentgen had produced a year earlier in experimentswith a gas discharge tube. Madame Curie called this phenomenon radioac-tivity. Further investigation by her and others showed that this propertyof emitting radiation is a specific property of the given element and thatcertain quantitative relationships exist. They also found that theradiations are continually emitted so that energy is constantly lost.These radiations, and therefore the energy, come from the nucleus of theatom. Also, the energy released by radioactive materials is about amillion times more intense than the energy released in standard chemicalreactions. Moreover, the atoms producing these radiations are unstable andemit radiation at characteristic rates to form new atoms. These, in turn,may transform to other substances, but the end product will be morestable.

By the use of magnetic fields, it was shown that there are threedistinct types of radiation (see Figure 2.1). These were given thedesignations: alpha (a>, beta (B> , and gamma (7) radiation.The firs t two types can be deflected in magnetic fields, indicating thatthey are charged particles. Gamma radiation cannot be deflected inmagnetic fields. It is not a charged particle but a form of electro-magnetic radiation, similar to light.

Since a particles were deflected in a different direction in themagnetic field than were j3 particles, the two were oppositely charged.From the direction in which the a particle was deflected, it was shownto be positively charged, whereas the /? particle is negativelycharged. Further investigation revealed that the a particle isactually a helium nucleus (>e - two positive charges, mass number 4)

and the /3 particle is actually an electron (e- - one negativecharge, mass of 0.000549 u).

erationalHealth

Physicsning (Moe)


2/32

ALPHA- GAMMA

RADIOACTIVESOURCE

Figure 2.1 Radiation deflection in a magnetic field. The field isperpendicular to and directed into the plane of thepaper. (R.E.Lapp / H.L.Andrews, NUCLEAR RADIATIONPHYSICS, 2/e 1954, 1~73. Reprinted by permission ofPrentice-Hall, Inc., Englewood Clif fs, NJ).

Following Becquerel's discovery, further studies revealed that threenaturally occurring radioactive series existed. The parent, or firstmember, of these series, or chains, are 232n, 9 238~ and 236~and the series are called the thorium, uranium and actinium series,respectively. These chains have certain common properties: each parent hasa very long half life (time for half the initial nuclei to decay - see2-D); each has a gaseous member which is a radon (Rn) isotope; and eachends up as a stable lead (Pb) isotope. An artificially occurring series,called the neptunium series, beginning with 241Pu but named after thelongest-lived radionuclide in the series, 237Np, was later produced byneutron bombardment of 238U. 1

With the discovery in 1934, that radioactivity can be induced in astable atom, a number of additional decay modes, have emerged (such aselectron capture and isomeric transitions). In addition, a number of newparticles, associated with radioactive decay, have also been uncovered.These are the neutrino and the anti-neutrino. These particles have beenuseful in the explanation of /3 decay. The neutrino and anti-neutrino

erationalHealth

Physicsning (Moe)


3/32

2-3are almost massless particles (< 10m3 of the electron mass) whichcarry away energy in decays. Also, the anti-particle to the electron, thepositron, was discovered, as well as radioactive decay in which positronsare emitted. In addition, the advent of high energy accelerators hasincreased the complexity of the number of known particles and their decaymechanisms.

Other naturally occurring radionuclides, such as 4OK > 147Smand 137Re, etc., which are not members of long series, or decaychains, have also been found.

B. Radioactive Transformations

When an atom undergoes a transition, or decays, by the emission ofan alpha particle, the atomic number Z decreases by 2 and the mass numberA decreases by 4, so that

This can be shown by a diagram called a decay scheme (see Figure2.2.a). In this diagram, different energy states are indicated by avertical scale. In the figure, the ground state of the parent atom isshown at a higher vertical position (or higher energy level) than theground state of the decay product (called the daughter product). In thedecay scheme layout, the emission of positively charged radiation is shownby an arrow to the left; negatively charged radiation by an arrow to theright. An example of the a decay is 233u 234m + 4He, inwhich 238U is the parent, 234Th is the daughter and He is thea particle. In the diagram shown, the a is emitted with a discreteenergy which equals the net energy difference between the energy level ofAX and that of A-4y ( accounting for the masses of the parent, daughterZ z-2and the a particle as well as the recoil energy of the daughter). Insome cases, the emission of the a will not bring the decaying nuclideto the ground state of the daughter, so that emission of a y ray mayfollow.

erationalHealth

Physicsning (Moe)


4/32

2-4

AXZ

a) Alpha Decay

Y

z-1

cl B++ Y Decay

A

YY A Yz 1

b) B+ Y Decay

d) EC Decay

e) IT Decay

Figure 2.2 Decay schemes for various modes of decay: aI $ $ y, EC and IT.

erationalHealth

Physicsning (Moe)


5/32

2-5

When the transformation is by negative j3 emission, the atomicnumber Z increases by 1, but the mass number A remains the same. The anti-neutrino i7 carries off excess energy. Often, a 7 ray is emittedfollowing a B- decay (see Figure 2.2.b). Negative j3 emission islikely to occur if the n/p ratio is too high. The decay is given by

AX + AZ zJl+ f3- + Yin which p- represents the electron, 'e. The emission of a gammaray (as shown in Figure 2.2b) does -' not affect either the atomicnumber or the mass number. It is indicated by a vertical straight line inthe figure.

Positron (/3+ or Oe) emission is shown1 in Figure 2.2~. Thereaction isAX +Az z- x +p++u

in which v is the neutrino that carries away excess energy. As in thecase of fi- decay, a -y ray may be emitted after positronemission. For positron emission to occur, the parent mass must exceed thedaughter mass bY more than two electron masses (2 moc2). Analternate mode, which often competes with positron decay is electroncapture (EC), pictured in Figure 2.2d. It should be noted that somediagrams picture /3+ decay in the same manner as electron capture. Ofcourse, if the condition for positron emission stated above is not met,then only electron capture may occur. For this reaction,

$rloe-tA +u.z- 1The electron which is captured is most often a K shell electron, althoughL and M capture are also possible. X rays which are emitted followingelectron capture will be those from element Y. Positron decay and/orelectron capture are likely to occur if the n/p ratio is too low.

erationalHealth

Physicsning (Moe)


6/32

2-6Sometimes a nucleus in an excited state will emit energy as a y

ray, rather than emit a particle. This is called an isomeric transition.As stated before, this emission does not affect the atomic number or themass number, so

The asterisk indicates that the nucleus is in an excited state. In Figure2.2e, the excited state is indicated by AmX, where the m stands formetastable. In this decay scheme, two paths are available. A photon may beemitted from the excited level to the ground state of AX, followed byB- emission. The alternate mode involves emission of a higher energyB- in the transition to AY. The branching ratio gives the fractionz+1

of the total transitions which proceed by each given mode.An alternate mode of decay for an excited nucleus is called internal

conversion. Instead of the emission of the 7, the nucleus transfersthis energy to one of the orbital electrons, which is then ejected with adiscrete energy (see 3.2).

Another mode of decay observed in high mass number substances(D230) is spontaneous fission. This process often competes with CYdecay. The reaction is

AX+A'y + A-A ' -kW + k 1,z Z' Z-Z' 0In the above, the original nuclide Ax splits into two fragments Y and W,and a number, k, of neutrons are emitted. The same number of neutrons willnot be released in each fission, so k is an average value. The totalnumber of n released in spontaneous fission varies with the mass and theatomic number. Nuclides with even Z and A seem to have the shorter spon-taneous fission half lives.2 Since fission releases neutrons, fragments,prompt and delayed -y rays and B particles, this complex mode issome times indicated by simply a vertical arrow marked SF.3

If a nuclide has enough energy in an excited state, it may alsodecay by neutron emission.4 Since the process itself is rare, and the

.I

erationalHealth

Physicsning (Moe)


7/32

2-7emission time very short, practical neutron sources, such as those fora, B 01: 7, are not readily found.

Included with the diagram of the decay scheme is other usefulinformation about the nature of the process. Sometimes, the complex natureof the process does not allow all of the information to be displayed.Shown in Figure 2.3, is the decay scheme for 24Na, adapted fromReference 5, to point out some of the data which is supplied for simpledecay schemes.

In the figure, 24Na is indicated as a /3- emitter with ahalf life of 15 h. For the two p- groups which are emitted, thefraction of transitions in which that p- is emitted (often calledthe intensity), as well as the maximum energy (in MeV) of the emitted/3 is shown. That is, fi-, is emitted in 99.9% of the transitions1and the maximum energy of this p is 1.39 MeV. The emission of the

4.123

.;014.144 MeV

\

Y2 1.369 MeV 1 o.o

;;Mg (STABLE)

Figure 2.3 Decay scheme for 24Na.

erationalHealth

Physicsning (Moe)


8/32

2-8/I produces 24Mg but in an excited state, either 4.123 MeV abovethe stable 24Mg ground state, or 1.369 MeV, if /I- is emitted.The excited nucleus returns to the ground state by 2 emission of yradiation. For the majority of decays (99.9%), two 7 rays, one of2.754 MeV immediately followed by one of 1.369 MeV, are emitted. For thesmall number of transitions involving p-2 (O.l%), only one 1.369 MeV7 is emitted.

Transformation or decay schemes can be found in diagram form inReferences 3,5-7, along with additional information concerning thetransition and other nuclear properties. Reference 3, which has beenprepared specifically for use in internal dosimetry calculations, givesthe average /3 transition energy, rather than the maximum energy. Othersources of useful decay scheme data, in tabular form rather than diagrams,can be found in References 8-10.

Another useful source of information about unstable (or radioactive)nuclides is the Chart of the Nuclides (Appendix G). This chart lists wellover 2,000 nuclides, of which only about 300 are stable. The generalfeatures of the chart, of interest at this point, are shown in theexcerpt, Figure 2.4. Each horizontal row represents an element, i.e.,H = hydrogen, He = helium, and the individual filled spaces in a row arethe known isotopes of that element. Note that hydrogen has three knownisotopes. The atomic number Z increases vertically with each row. Theneutron number N = A-Z increases horizontally. The number of each columnis the total number of neutrons (n) for each nucleus in that column.

The heavily bordered space at the far left side of each row gives thechemical atomic weight (in atomic mass units) of the element as found innature (a combination of lH and *H). Shaded spaces in the row arestable isotopes. A black rectangle at the top of the space indicates aradioactive isotope that is found in nature. White spaces represent radio-nuclides which are artificially produced.

For stable nuclides, such as 1H, the firs t line of the square givesthe symbol and the mass number. The second line gives the isotopicabundance. This is the percent of 1H which is present in the element

erationalHealth

Physicsning (Moe)


9/32

2-9

He4.0026

1.00797

nlll.Om

p-O.7821.008665

0 1 2

NFigure 2.4 Excerpt from Chart of Nuclides. (Knolls Atomic PowerLaboratory. Schenectady, New York. Operated by theGeneral Electric Co. for Naval Reactors, The U. S.Department of Energy).

hydrogen as it is found in nature. The number at the bottom of the spaceis the atomic mass of the neutral atom lH.

For the unstable nuclides, such as 3H, the first line gives thesymbol and mass number. The second line gives the half life. Additionallines present decay modes and energies (in MeV). For 'H, the listingindicates a /3- emitter of maximum energy 0.0186 MeV which is notfollowed by any 7 emission. In the case of very long-lived, naturallyoccurring radionuclides, the second line gives the abundance. Subsequentlines give the data for the decay modes.

For certain radionuclides, a long-lived excited nuclear state ispossible. These are called isomeric states. Each state has its own halflife, decay mode, and energy of emissions. Isomers are shown on the chartby a divided square for the given radionuclide.

To trace the product of a radioactive decay, or to find a productnucleus when a target is bombarded, the scheme in Figure 2.5 may be used.It is designed to allow easy location on the Chart of the Nuclides. Forexample, let the original nucleus have an atomic number Z and neutronnumber N, and decay by a emission. The product nucleus will be found

erationalHealth

Physicsning (Moe)


10/32

2-10z+2

z+1 P- pout in

He ain in2Hin $HinZ n oflK;Iw nout NUCLEUS in+z-1 3H 2H P JsECout out out out

az-2 3Heout outN-2 N-l N N+l N+2

Figure 2.5 Location chart for nuclear products. (Knolls Atomic powerLaboratory, Schnectady, New York. Operated by theGeneralElectric Co. for Naval Reactors,the U&Departmentof Energy.)

on the chart at the location Z-2 and N-2. In the case of 238U which isan a emitter, Z=92 and N=146, so the daughter product will be found atZ=90 and N=144. From the Chart in Appendix G, this location corresponds to234Th. For tracing a radioactive chain, the first daughter can then beconsidered the original nucleus to find the second daughter, and so on.

c. Decav LawWhen one deals with large numbers of atoms, it is found that all

radioactive substances follow the same general decay pattern. For example,assume one has a radioactive source and some means of counting the numberof atoms which decay in a given time interval. The number of atomsdecaying in a given time interval is called the activity of the sample. Ifone plots the percentage ratio of the activity at some later time (t) tothe activity at time t=O, on linear graph paper (Figure 2.6) versus thetime t, the curve obtained indicates an exponential or logarithmic rela-tionship. If the same ratio is plotted on semilog paper (Figure 2.7), astraight line is obtained. This indicates that radioactive decay is an

erationalHealth

Physicsning (Moe)


11/32

80

60

1

Figure 2.6100

;;z5 10i=2

1

\

2 3 4 5 6

TIME UNITSRadioactive Decay, linear plot.

\

\

\

0 1 2 3 4 5 6

TIME UNITSFigure 2.7 Radioactive decay, semllog plot.

erationalHealth

Physicsning (Moe)


12/32

- _-- .-_- __ ----

2-12exponential (logarithmic) process, that is, there is a constant fractionaldecrease in the decay rate during equal units of time. Although the samenumber of atoms do not disintegrate during each unit of time, the samefraction of the atoms present decay during a unit of time. The disinte-gration rate is proportional to the number of atoms present:

N -NAN ' o aN=- ,At Vto2.1

where N is the number of atoms present at any time t. By including a con-stant of proportionality, the expression becomes

AN-= -J,N 3 2.2At

where X, called the transformation constant (also the decay constant),is the constant of proportionality and the minus sign indicates a decreasein the decay rate as time increases. When the expression is integrated, weget the general exponential relationship for radioactive decay:

N = NoewXt 2.3where N is the number of atoms left at time t from a sample of N0radioactive atoms initially present.

If we take the logarithm of each side of this equation, we getIn N = In No-Xt. 2.4This is similar to the slope-intercept form of a straight line, i.e.,y= a + bx. Thus, if one uses semilog paper and plots the values of N onthe logarithmic scale versus the time on the linear scale, the slope ofthe resulting straight line will be -), and the intercept (the value ofN at t = 0) will be No.

An an example, consider the data:

Number of Atoms Time (s)80,000 029,432 6010,824 1203,984 1801,464 240536 300200 360

erationalHealth

Physicsning (Moe)


13/32

2-13Plot a curve on semilog and on linear paper, and determine the decayconstant for the radioactive element.

To plot on semilog paper: Take three-cycle semilog paper (this willcover the ranges 100-1000, lOOO-10,000 and lO,OOO-100,000) and plot thenumber of atoms on the logarithmic scale and the corresponding value ofthe time on the linear scale. The plot will be a straight line.

To find the slope of the line (-X): Take the ratio of the numberof atoms for two times; find the natural log of this ratio, and dividethis log by the difference in time for the values chosen. For example,choose the readings for t = 0 and t = 120 s. Take the ratio of the numberof atoms at the later time to the number of atoms at the earlier time.This ratio is 0.1353. The natural log of this ratio is In 0.1353 = In(1.353 x 10-l) = In 1.353 + In 10-l = ln 1 353 _ ln 10 =.0.3026 - 2.3026 = -2. The time difference for the two values is 120 s. Wenow obtain

lnN=lnN, - Xt;In N - In No = -Xt;In N/No = -Xt;-2 = -X(120 s);1.67 x 1O-2 s-l = x .

To plot on linear paper: Plot number of atoms as ordinate (y axis)and the time as abscissa (x axis).

The decay constant X expresses the probability that a singleatom will decay in a unit of time. The larger the value of X, the morerapidly a radioelement decays.

One is usually more interested in the decay rate of a given samplerather than the number of atoms present. The activity of a sample isexpressed as:

a = At = XN.dt 2.5

erationalHealth

Physicsning (Moe)


14/32

2-14

In other words, the activity At of the sample can be found bymultiplying N, the number of atoms present at any time, by the decayconstant X of the element. Accordingly:

N = NoeeXt;AN = XNoemXt;At = AoemXt, 2.6

where A, is the activity (decay rate) of the sample at t = 0. Combiningequation 1.2 and 2.5 above, the expression for the activity of a knownmass m of a radionuclide is given by:

At = XN = p N,2.7

D. Half Life and Mean Life

The decay constant X is closely related to the concept of thehalf life of a radionuclide. The half life TG is defined as the timerequired for the activity of a given nuclide to decay to one-half of itsinitial value. Consider an initial activity A,. At some time t = T%,the activity will be equal to A,/2. From our general relationship, then,

A, = AoemXt;

IA =Ae -XT%2O O

1=e -XT%2

In 1 - In 2 = -XT,,XT% = In 2;

TG = kiLti 0.693,x x

2.8

. .

erationalHealth

Physicsning (Moe)


15/32

2-15substituting for X in equation 2.7 gives an alternate expression forthe activity in terms of the half life,

XmNa=At= A 0.693 m NaAT% 2.9Example: Find the activity (dis/s) of a 10-3 kg (1 g) sample ofsCo (T% = 5.27 y).In SI units, a mole of 6oCo = 0.06 kg, and

-3 23 moleculesAt

= 0.693 (10 kg) 6.022~10 mole= 4 19x1013dis/s0.06 > 5.27 y 3.1536~10~~mole Y

The actual length of time any one atom survives may be anything from 0 toinfinite time, theoretically. Suppose we were able to sum up all thelifetimes for the entire sample of atoms. Now, divide by the total numberof atoms, and we arrive at the average lifetime of an atom. The average,or mean life, T is given by:

T1 T%z-=-iix In 2 1.443 T,, 2.10

Given the activity At of a sample, the total number of transitions whichwill occur in the sample may be obtained from:

Trans. = AtT = 1.443 Ttit 2.11

in which the time units must be consistent. For the 6oCo exampleabove, the total number of transitions would be:

Trans. = 1.433(4.19x1013~)(5.27 y)(3.1536x107 'Y'= 1.005x1022.

erationalHealth

Physicsning (Moe)


16/32

2-16Each radioactive atom has its own unique pattern of decay. Three

aspects which characterize the radioactive decay of a given nuclide are:

1) half life of emission,2) energy of the emission, and3) type of emission

The identification of a particular radionuclide will normally dependupon how well one can determine these three factors. Many radionuclideshave half lives which are nearly the same, but the energy of theiremissions differs greatly. On the other hand, many radionuclides havesimilar energy of emission, but their half lives differ greatly.Oftentimes, a number of techniques or combination of techniques willbe required for the identification of an unknown radionuclide. Sometimes,when one knows the types o f radioactive material being used in an area, itis possible to pinpoint the identity of the radionuclide by the half lifealone. For long-lived radionuclides, however, this approach may not befeasible. Also, for faster identification, it may be easier to analyze forthe energy of the emissions or to search for the type of emission.

E. Activity Units

The SI unit for the activity of a radioactive substance is thebecquerel (Bq). A becquerel is the activity of a radionuclide whose trans-formation rate is 1 dis/s. This unit replaces the curie (Ci), thehistorical activity unit originally taken as the decay rate of 1 g of226Ra, but later defined as a transformation rate of 3.7~10~~dis/s.4 The two units are related by

1 Bq = 2.703~10-~~ Ci. (1 Ci = 3.7~10~~ Bq)

The becquerel is a rather small unit, whereas the curie is a rather largeunit. To aid in designating the large range of values that may beexperienced, SI prefixes are shown in Table 2.1 (adapted from Reference11).

erationalHealth

Physicsning (Moe)


17/32

2-17

Table 2.1 - SI Prefixes

Prefix Symbol Factor Prefix Symbol Factordekahectokilomega&aterapetaexa

dahkMGTPE

101 deci102 centi10s milli106 micro109 nano1012 pica1016 femto1018 atto

d 10-1C 10-zm 10-eP 10-an 10-eP 10-12f 10-15a lo-18

Common smaller units of the curie encountered in health physics workare the pCi, nCi and $i. In terms of SI units, these become

1 pCi = lo-l2 Ci = 3.7~10~~ Bq = 37 mBq1 nCi = 10eQ Ci = 3.7~10~ Bq = 37 Bq1 PCi = 10m6 Ci = 3.7~10~ Bq = 37 kBq

Some laboratory sources in use will be more conveniently expressed in MBq.The definition of activity refers to the transformations (disinte-

grations) per unit time. Oftentimes, the number of transformations willdiffer greatly from the number of radiations emitted, as for example, inthe case of 6Oco. For each 6oCo atom which decays, a beta particleand two y rays are emitted. In this case, then 1 MBq of (jCo wouldemit 1.0x106 8/s and 2.0~10~ y/s. If the activity of a sampleis to be accurately calculated from experimental data, the decay scheme ofthe radionuclide must be taken into account.

If counting a p- emitter, some of the 7 rays may also becounted. If one merely takes the counting rate and converts this to atransformation rate, without considering the decay scheme, one willusually overestimate the activity of the source. For example, consider acounter which counts all the fi (1 count/dis) and 2% of the gammas. Incounting a 1 MBq 6oCo source, the observed counting rate would be1.0x106 counts/s p + 0.02 (2.0x106) counts/s 7 = 1.04x106 counts/s. If one converts this activity to MBq without regard to

erationalHealth

Physicsning (Moe)


18/32

2-18the decay scheme, the activity would come out as 1.04 MBq, a slightoverestimate. However, if one counted this same source in a counter, whichdid not count the p, a serious overestimate would occur. Suppose sucha counter counted all of the 7 rays (1 count/dis). Then, the resultwould be 2x lo6 cts/s, which converted to activity would be 2 MBq,double the true source activity.

In the case of electron capture decay, a serious underestimate couldoccur when counting, say 7Be, in such a counter. 7Be emits about0.1 7 per transformation. This means a 1 MBq source will only giveabout 1x105 cts/s , which for this counter could be erroneouslyinterpreted as 0.1 MBq.

In counting an a emitter, the situation is usually different.Since a proportional counter is normally used, the 7 rays associatedwith an Q emitter would not be counted, and thus the activity can bedetermined from the counting rate. Nevertheless, if the source beingcounted contains a radioactive chain, there may be more than one Qemitted per transformation, since daughter products may also emit Qparticles. This would subject one to the potential of overestimating thesource activity.

When the radionuclide that one is counting is known, a proper choiceof counting equipment can be made. A suitable calibration of thisequipment will then allow one to obtain a reasonable activity estimate.When the radionuclide is unknown, as occurs frequently in health physicswork, one must exercise care in the interpretation of any counting data.

F. Snecific Activity

The specific activity is defined as the activity per unit mass of asubstance. It has usually been expressed as Ci/g of the radioelementbefore SI units. The shorter the half life of the emitter, the greater isfound to be its specific activity. The specific activity can be calculatedfrom equation 2.5, now expressed as

SP.A. = XN. 2.12

erationalHealth

Physicsning (Moe)


19/32

2-19where X is the transformation constant and N is now the number ofatoms in one kg of the radioelement. 6oCo has a half life of 5.27 y.

Calculate the specific activity:SP . A. = XN=0.693i&NT% Aa' and A = .060 moles

0 693 (1) 6.022~10 23a5.27 y (3.1536~10~ ;)*06 = 4.19x1016s~

[SP.A.-4.19x10 l6 hiIkg (2.703~10 -lL1 31 -3 k = 1.13~10~ Ci/g]gTo convert a tabulated value of SP.A. in Ci/g to SI units, use

SP.A (Bq/kg) = 3.7~10~~ SP.A (Ci/g)

G. Decay ChainsIn general, most radioactive substances do not decay to form a

stable nuclide, that is to say, the daughter nucleus is also radioactiveand decays with its own characteristic half life. The problem ofdetermining the amount of the daughter present at any time depends,therefore, upon both half lives. The daughter will be produced at acertain rate from the parent, but will decay with its own rate.

Suppose that at a given time there are NY parent atoms, with decayconstant %' and no daughter atoms. After a certain interval oftime, At, the increase in the number of daughter atoms, AN2, isgiven by

AN2 = (decay rate of parent - decay rate of daughter) At. 2.13The decay rate of the parent is actually the formation rate of thedaughter, since whenever a parent atom decays, it becomes a daughter atom.The decay rate of the parent is XINl, where Nl is the number of

erationalHealth

Physicsning (Moe)


20/32

2-20parent atoms present at any time. Similarly, the decay rate of thedaughter is X2N2. The expressiondaughter atoms per unit time becomes

for the rate of change of

AN2= XINl- X2N2.AtFrom the general relationship,

time isNl = NyemXlt.

2.14

the number of parent atoms at any

2.15Substituting this into the above expression and integrating the equation,one arrives at

0NIXl (e-Xlt-e-X2t).N2=,2-X1 2.16

where N2 is the number of daughter atoms present at any time.When the parent half life is long compared with that of the

daughter, then y$, and the term e-X2t becomesnegligible compared with eWXlt after a suffic iently long time.Then equation 2.16 reduces to

However, since

Nl = NieeXlt.we find

N2 =x 1Nlx2 -Xl'

2.17

2.18A condition is thus reached in which the ratio N2/Nl remains constant.This state is called transient equilibrium. In this case, the daughteractivity decays at the same rate as the parent activity. Another way ofstating this is that the formation rate of daughter atoms equals the decay

erationalHealth

Physicsning (Moe)


21/32

2-21rate of the daughter atoms. In each unitfractional decrease in the parent andconstant ratio.

of time then, there is the samedaughter activities, yielding a

When the parent activity is extremely long-lived, as is the casewith 238U (TG=4.5x10g Y) f then XlX2. The daughter rises to a maximum and decays with its owncharacteristic half life. In this case, no equilibrium is reached.

The above relationships have been presented for cases in which nodaughter atoms are initially present. If there are daughter atoms, we mayadd a term N2e -X2t to equation 2.16, which gives

N2 VP- (emXlt -emAx2-x1 )+ Ni e-X2t 2.210in which N2 is the number of daughter atoms present at t = 0. In most

instances, one is interested in the activity of the sample, rather thanthe number of atoms. The activity can be obtained by use of equation 2.5,

At = JVX = N2X2 for this case,(AtI2 = N2X2

0= X2A1 (e-Xlt-e-X2t)+A2e-X2tcy+1>

erationalHealth

Physicsning (Moe)


22/32

2-22When a radioactive chain is present, the method outlined above can

be extended. The original relationships were developed by H. Bateman in1910. A general equation for the quantity of the n& member of a seriescan be found in Reference 12.

H. Decav Curve of a Mixture

The problem of identifying the half life of a radionuclide from adecay curve can be complicated by the fact that the experimental curve maybe a composite of the contributions from a mixture of radionuclides. Whenthe activity values are plotted on semilog paper, the curve obtained fromsuch a mixture will contain: 1) an initial straight portion, followed by2) a curved portion, which is then followed by 3) a final straightportion. This type of decay curve is explained by the presence of at leasttwo radioactive products in the sample being counted.

The initial straight portion of the curve represents the sum of theactivities of all the components, whereas the final straight portion isdue only to the activity of the longest-lived component of the mixture.The shorter-lived products contribute greatly to the decay rate initially;but as they decay out, the total activity will not decrease as rapidly asin the beginning. This accounts for the curvature in the plot.

Since the final straight portion represents the activity of thelong-lived component, this line can be extrapolated back to zero time. Theinitial activity, of the long-lived component, is the intercept value att = 0, of the extrapolated line. In addition, this line gives the decaycurve of the long-lived component. When a series of values along thiscurve are subtracted from corresponding values on the original curve, theremainders will represent the activity of the short-lived components, atthe corresponding times. If the mixture contains only two components, thena plot of the remainder values will give a straight line. As before, theactivity value at t = 0 for this new straight line will represent theinitial activity of the short-lived component.

The half life for each component can then be determined graphicallyfrom each of the straight lines.

erationalHealth

Physicsning (Moe)


23/32

2-23If three or more components are present, this method can be extended

provided the original data are sufficiently accurate.Example: Using the data in the table, plot a curve on semilog (see

Figure 2.8) and determine the. half lives for all the activities in thesample.P1; ZOO5;7884,0783,1482,5161,6661,125770

531

t 01234681012

GE t370 14258 16181 18127 2089 2262 2444 2631 28

Plot the data on 3-cycle semilog paper. From the shape of the curveone would surmise that there are at least two components. Extrapolate thefinal straight portion of the curve back to zero time. This line inter-sects the t = 0 axis at 4000 cps. The half life is determined directlyfrom the graph: find the time at which the activity has dropped to 2000cps. This occurs at 4 h. The half life can be determined mathematically bychoosing two points on the curve, finding the ratio of these two points,and dividing the natural log of this ratio by the difference in timebetween these two points. This gives the value of A. Take the points:t = 0, cps = 4000; t = 12 h, cps = 500.

1 = ln$ = -In 8 = -12 A;0

x In 8 3 In 2 In 2 h-lZ-G =-12 12 4

ButIn 2T%=-T-= E(4) h = 4 h.

Take points on the extrapolated curve and subtract their values fromthe corresponding points on the experimental curve. Plot these differences

erationalHealth

Physicsning (Moe)


24/32

2-24

C CURVE OBTAINED 8Y SUBTRACTING B FROM AAPOLATED CURVE OF THE 2-HOL-LIFE ACTIVITY

OBTAINED EIY SUBTRACTING D

JR

FROM C

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28Time (Hours)

Figure 2.8 Decay curve using example data in text.

erationalHealth

Physicsning (Moe)


25/32

2-25

versus the corresponding times. Consider the points for t = 0; the valueon the experimental curve is 10,000 cps, the value on the extrapolatedcurve is 4,000 cps. The difference, 6,000 cps, is plotted for t = 0.Similarly, for the point t = 1 h, the value on the experimental curve =5,778 cps, and the value on the extrapolated curve = 3,400 cps. Thedifference, 2,378 cps, is plotted for t = 1 h. Proceeding in this way, onecan determine enough points to draw the curve.

In this case the curve is not a straight line, it is a curved line.Once again extrapolating the final straight portion back to t = 0 gives aninitial activity of 2,000 cps. The half life of this activity asdetermined from the graph is 2 h.

When points on this extrapolated curve are subtracted from thecorresponding points on the second curved line and the differences areplotted, a straight line is again obtained. The initial activity of thiscomponent is 4,000 cps, and its half life as determined from the graph is30 minutes.

1.

2.

3.

4.

5.

6.

7.

8.

REFERENCESGlasstone, S., SOURCE BOOK ON ATOMIC ENERGY, 3rd ed, D. Van NostrandCo., Inc., Princeton, NJ (1967).Denham, D.H., Health Physics Considerations in Processing Trans-Plutonium Elements, Health Phys. 16, 475-487 (1969).ICRP Publication 38, RADIONUCLIDE TRANSFORMATIONS, Annals of theICRP, Pergamon Press, Oxford, England (1983).Knoll, Glenn F., RADIATION DETECTION AND MEASUREMENT, John Wiley andSons, New York, NY (1979).Lederer, C.M. and Shirley, V.S., Editors, TABLE OF ISOTOPES, 7th ed,John Wiley and Sons, Inc., New York, NY (1978).Shleien, B. and Terpilak, M.S., Editors, THE HEALTH PHYSICS ANDRADIOLOGICAL HEALTH HANDBOOK, Nuclear Lectern Assoc., Inc., Olney,MD (1984).Brodsky, A.B., Editor, HANDBOOK OF RADIATION MEASUREMENT AND PRO-TECTION, Vol. 1, CRC Press, West Palm Beach, .FL (1978).NCRP Report No. 58, A HANDBOOK OF RADIOACTIVITY MEASUREMENTS PRO-CEDURES, NCRP, Bethesda, MD (1978).

erationalHealth

Physicsning (Moe)


26/32

9.

10.

11.

12.

2-26

Kocher, D.C., A HANDBOOK OF DECAY DATA FOR APPLICATION TO RADIATIONDOSIMETRY AND RADIOLOGICAL ASSESSMENTS, U.S. DOE Report DOE/TIC-11026 (1981).Erdtmann, G. and Soyka, W., THE GAMMA RAYS OF THE RADIONUCLIDES,Verlag Chemie, Weinheim, NY (1979).NCRP Report No. 82, SI UNITS IN RADIATION PROTECTION AND MEASURE-MENTS, NCRP, Bethesda, MD (1985).Skrable, K.W., et al, A General Equation for the Kinetics of LinearFirst Order Phenomena and Suggested Applications, Health Phys. 27,155-157 (1974).

BIBLIOGRAPHYKathren, R.L., Historical Development of Radiation Measurement and Pro-tection, HANDBOOK OF RADIATION MEASUREMENT AND PROTECTION, Ed. by A.Brodsky, CRC Press, West Palm Beach, FL (1978).Hendee, W.R., MEDICAL RADIATION PHYSICS, 2nd ed, Year Book MedicalPublishers, Chicago, IL (1979).Cember, H., INTRODUCTION TO HEALTH PHYSICS, 2nd ed., Pergamon Press,Oxford, England (1983).Lapp, R.E. and Andrews, H.L., NUCLEAR RADIATION PHYSICS, 4th ed, PrenticeHall, Inc., Englewood Cliffs, NJ (1972).Evans, R.D., Engineers' Guide to the Elementary Behavior of RadonDaughters, Health Phys. l7, 229-252 (1969).Friedlander, G., et al, NUCLEAR AND RADIOCHEMISTRY, 2nd ed, John Wiley andSons, New York, NY (1964).Glasstone, S., Chapter 5, SOURCEBOOK ON ATOMIC ENERGY, 3rd ed, D. VanNostrand Co., Inc., Princeton, NJ (1967).Johns, H.E. and Cunningham, J.R., THE PHYSICS OF RADIOLOGY, 4th ed,Charles C. Thomas, Springfield, IL (1983).Rees, D.J., HEALTH PHYSICS, Massachusetts Institute of Technology Press,Cambridge, MA (1967).Caro, D.E., et al, INTRODUCTION TO ATOMIC AND NUCLEAR PHYSICS, AldinePublishing Co., Chicago, IL (1962).Kathren, R.L., RADIOACTIVITY IN THE ENVIRONMENT: SOURCES, DISTRIBUTION,AND SURVEILLANCE, Harwood Academic Publishers, New York, NY (1984).

erationalHealth

Physicsning (Moe)


27/32

2-27/'Shapiro, J., RADIATION PROTECTION, 2nd ed, Harvard University Press,Cambridge, MA (1981).

Gollnick, D.A., BASIC RADIATION PROTECTION TECHNOLOGY, Pacific RadiationPress, Temple City, CA (1983).Hurst, G.S. and Turner, J.E., ELEMENTARY RADIATION PHYSICS, Wiley andSons, New York, NY (1969).

2.12.22.32.42.5

When and by whom was radioactivity discovered?Name the three distinct types of naturally occurring radiation.With what nucleus are Q particles identical?With what atomic particle are t9 particles identical?When an Q particle is emitted, how many nuclear particles areemitted?

2.6 How are the A and Z numbers of a radioactive atom affected when:ba: an a particle is emitted?a p particle is emitted?

2.7 When a p particle is emitted,a what nuclear particle was converted to release the Bparticle?b. what is the new nuclear particle that results from the con-version?

2.8 How does the emission of a 7 ray affect the atomic number andthe mass number?2.9

2.102.112.12

What is the relationship between the number of atoms present in aradioactive source at any particular time t to the time t?To what value is the disintegration rate of radioactive atoms pro-portional?What does the symbol X (lambda) represent in radioactive decayschemes?Describe the plot of radioactive decay rate versus time:

OUESTIONS

a> on linear (common) graph paper and,b) on semilogarithmic paper.

erationalHealth

Physicsning (Moe)


28/32

--.- . -.----. . _

2-28

2.13 How many cycles should semilogarithmic paper have to plot4 700 to 1300 atoms?b) 20 to 95 atoms?c> 300 to 120,000 atoms?

2.14 What is the base of4 common logarithms?b) natural logarithms?

2.15 What are the common logarithms ofa> 1c) lo2 b) 10010-4+ d) 1O-3e>

2.16 What are the natural logarithms ofa> :-1v b) e3c> .

2.17 What value results when the number of radioactive atoms present (N)is multiplied by the decay constant (A)?2.18 What value is defined by the time required for a given isotope todecay to one-half its original value?2.19 If a radioactive substance loses half its activity in 3 hours, what

will its half life be 6 hours later?2.20 In the conversion formula

a> what does the symbol TQ represent?b) what does the symbol X represent?c> of what number is 0.693 the natural logarithm?2.21 What is the unit of source activity based upon the number of disin-tegrations per unit of time?2.22 Why may j? emitters appear to have a higher becquerel value thantheir true value?2.23 What does the activity of one kilogram of a radioactive substancedefine?

erationalHealth

Physicsning (Moe)


29/32

2-29

2.24 What terms identifya> a radioactive material that, upon decay, results in anotherradioactive material?b) the resulting radioactive material?

2.25 Under what condition(s), does the ratio of the number of daughteratoms to the parent atoms remain constant ? What is the name given tosuch a state?2.26 Under what conditions is secular equilibrium reached between thedaughter and parent?2.27 Under what condition is no equilibrium reached?2.28 Assuming that good data has been obtained, what does a curved decay

plot on semilogarithmic paper indicate?2.29 What term is given to the projection of a mathematical curve beyondknown data?2.30 What value is obtained by taking natural logarithm of the ratio oftwo points on a radioactive decay curve and dividing that ratio bythe intervening time?

PROBLEMS

2.1 Complete the following decay schemesa. 'fiRa+Rn + He b) l46 C+N + /3- + i

C. 1;~g- 107? 7 d) Au+';%g + ,8- + L

2.2 If the number of radioactive atoms at time t is 2x106, and 2x104 atoms disintegrate in 5 minutes, what is the approximateradioactive constant?Answer: - 2x10m3 minutes

2.3 If the radioactive constant is O.l/day and there are 3~10~ radio-active atoms, approximately how many atoms will disintegrate in 1minute?Answer: 208/min

erationalHealth

Physicsning (Moe)


30/32

2-30

2.4 From the formula At = A, emXt, find the activity of a sampleat 4:00 P.M. when its activity wa s 1000 disintegrations per minute at1O:OO A.M. The decay constant X of the sample is 0.2/day.Answer: 951/min

2.5 The half-life of radon is 3.8235 days. What is the decay constant?Answer: 0.1813 d-l

2.6 In problem 2.5, what percentage of a freshly separated sample of radonwill disintegrate in 1 day? In 2, 3, 4, 5, 10 and 20 days?Answer:

Time(d) % of radondecayed1 16.62 30.43 41.94 51.65 59.610 83.720 97.3

2.7 The activity of 10m7 kg of 230Th is found to be 7.2~10~Bq (dis/s). What is the half life of 230Th?90

Answer: 8.0~10~ years2.8 The activity of a radioactive sample is 25 Bq. What wa s the activity ofthe sample 1 hour earlier if the half life is 25 minutes?

Answer: 132 Bq2.9 Carbon-14, 14C, has a half life of 5730 years. What is the

specific activity of '%C?

Answer: 1.65X1014 Bq/kg

erationalHealth

Physicsning (Moe)


31/32

2-312.10 Krypton-88, 88Kr has a half life of 2.8 hours and its daughterrubidium-88, 88Rb has a half life of 18 minutes. If the krypton-88has decayed to 5xlOlO atoms over a period of several weeks, howmany daughter atoms are present? Hint: use the short form formula since

krypton-88 has a much longer half life compared with the half life ofrubidium-88.Answer: 6~10~ atoms

2.11 Thorium-232, 232m has a half life of 1.41~10~~ years anddaughter radium-22i, 228Ra has a half life of 5.76 years. If 10%atoms of thorium-232 are found in a lump of natural ore, how many atomsof radium-228 should be present? Hint: thorium-232 has an extremelylong half life when compared with the half life of radium-228.Answer: 4.085~10~ atoms.

2.12 The first two members of a radioactive series have half lives of 6minutes and 12 minutes respectively, while the third member is stable.Starting with lo8 atoms of the first member and none of the secondand third, plot the number of atoms of the three members as a functionof time. Determine from the graph (or otherwise), the time at which thesecond member reaches its maximum.Answer: 12 minutes

2.13 A smear obtained from a graphite block bombarded by a 50 MeV protonbeam was counted with a germanium detector. The following data wasrecordedTime (Minutes) Elapsed Counts Per Minute

5 840010 710415 600820 510225 430030 360040 259150 182060 131070 93280 661100 330

Plot the data as (a) linear plot and as a (b) semilog plot. Determinethe half life. Can you guess what could be the radionuclide?

erationalHealth

Physicsning (Moe)


32/32

_ -..... . - __.---_ ----_-- ----- -

2-32

2.14 The half life of 238U is 4.4683~10~ years. How many kilogramsof 238U is needed for an activity of 3.7~10~~ Bq (this is 1 Ciin the old system of units)?Answer: 2973 kg

2.15 The data in the following table represents the decay curve of a mixtureof two radionuclidesa> Plot the data on semilog paperb) Determine the half lives of the two radionuclides graphically.

(min) cpm (min) cnm0 60,000 40 14065 34,000 45 95310 20,000 50 66415 12,000 55 45720 7,500 60 32325 4,750 65 22230 3,140 70 15235 2,062 75 111

Answer: - 9.5 min and - 4.5 min

erationalHealth

Physicsning (Moe)

moe section 2

Documents