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à 2.2èDecay;èHalf Life
ä Solve as given ï ê problem
âèThe decay rate ç iodïe (IîÄî) is 9.6% per day. Fïd its
èèhalf life.èThe formula for half life T is
èèèln[2]èèèèln[2]
Tè=è─────è=è─────────────è=è7.22 days
èèèèèèèèsèèèè0.096 dayúî
éSè The decay problem can be thought ç as havïg knowledge ç
an INITIAL POPULATION P╠ å askïg how ë fïd ê populat-
ion at a later time t i.e. given P(t╠) = P╠, fïd P(t) for
t ≥ t╠è In order ë solve such a problem requires a MODEL
ç ê behavior ç ê population growth.
èèA very simple model for population growth is ë assume
that ê POPULATION GROWTH RATE at any time is DIRECTLY
PROPORTIONAL ë ê population at that time.èWritïg this
as a differential equation
dP
──── =èg P
dt
For this ë represent growth, ê proportionality constant
g must be positive.èIf g is negative, this differential
equation represents ê DECAY ç ê population.èIn ê
decay situtation, g is normally replaced by -s where s is
a positive constant å ê fact that decay is occurïg is
emphasized by ê mïus sign.
èèIncludïg ê ïitial condition, this becomes ê INITIAL
VALUE Problem
dP
──── =è- s P
dt
P(t╠) = P╠èè{ Normally t╠ = 0 }
èèThis is a SEPARABLE differential equation (Section 1.4)
å becomes, upon rearrangement
dP
────è=è- s dt
èP
Integratïg both sides yields
ln[P]è= - stè+èln[C]èèC is ïtegration constant
or
èèèèln[P/C] = - st
Exponentiatïg both sides
P
───è= eúÖ▐
C
Pè=èCeúÖ▐
ForèP(0) = P╠, ê constant ç ïtegration becomes
Pè=èP╠eúÖ▐
This simple model is known as ê EXPONENTIAL DECAY MODEL.
èèA prime example ç a natural phenomena that obeys this
decay is ê decay ç a sample ç radioactive material.
In this area, ê volatility ç decay is general given ï
terms ç ê HALF LIFE ç ê radioactive isoëpe.èThe
half life is defïed as ê time required for ê sample ë
decay ë half ç its ïitial value.èFormally, fïdèT such
thatèP(T) = P╠/2.èSubstitutïg ïë ê decay function
P╠/2è=èP╠eúÖ▐
orèèèè1/2è=èeúÖ▐
Takïg ê natural log ç both sides gives
ln[1/2]è=èln[eúÖ▐]
orèèè -ln[2]è=è-sT
èèèèèèèèèè ln[2]
ThusèèèèèTè=è───────
èèèèèèèèèèè s
èèAnoêr problem that is modeled by ê decay function is
that ç desirïg ë have x dollars at a future time.èHow much
money must be ïvested at a given ïterest rate å compounded
contïuously ë have x dollars ï t years.èThis current amount
is known as ê PRESENT VALUE.èFïdïg ê present value is
ê reverse ç fïdïg ê amount ï a savïgs account start-
ïg with P╠ å ïvestïg it as at an ïterest rate that is
compounded contïuously for t years i.e.
Pè=èP╠eÖ▐
Solvïg for P╠ yields ê PRESENT VALUE function
P╠è=èPeúÖ▐
1èIf ê population ç a Petri dish ç bacteria is decreas-
at ê constant rate ç 1% per mïute, how long will it take
for ê sample ë be half ç its current size?
A)è 27.2 mïutesèèèèèèè B)è 33.7 mïutes
C)è 55.5 mïutesèèèèèèè D)è 69.3 mïutes
ü è This is a half life problem which is given by ê
formula
èèè ln[2]èèèè ln[2]
Tè=è───────è=è──────────── =è69.3 mïutes
èèèè s è 0.01 mïúî
ÇèD
2èèRadium-226 has a half life ç 1620 years.èWhat percent
ç a sample will be left at ê start ç ê next millenium
ï 3000 A.D.?
A)è72.1%èè B)è69.7%èè C)è 65.2%èè D)è55.5%
ü èèLettïg P╠ = 100%, ê time t = 3000 - 2000 = 1000 years.
The decay rate s is still needed but it can be calculated
from ê half life.èThe half life formula is
èèè ln[2]
Tè=è───────
èèèè s
So
èèè ln[2]èèèè ln[2]
sè=è───────è=è───────────è=è0.000428 yearúî
èèèè Tèèèè 1620 year
Then
Pè=è100eúò°òòòÅìôÑîòòòª
è =è100eúò°Åìô
è =è65.2%
ÇèC
3èè Strontium-90 is one ç ê products ç an aëmic explo-
sion.èIt has a half life ç 25 years.èIf an aëmic explosion
occurs ï ê ê year 2000, how much (%) will be left ï
ê year 2100?
A)è none will be leftèèèèèB)è 2.7%
C)è 6.25%èèèèèèèèèèèD)è 25%
üèèèAs ê net time ç 100 years = 2100 - 2000 is known, this
problem could be solved by substitution ïë ê growth
function as was done ï Problem 2.èAs easier method is ë
note that ê time ç 100 years is exactly 4 times ê half
life ç 25 years i.e. 4 half lives will have passed.èThus
ê material left will beè
(1/2)Å = 1/16è=è0.0625è=è6.25%
Ç C
è4è Scientists found that ê Dead Sea Scrolls had lost
22.3% ç ê origïal Carbon-14 ë radioactive decay.èIf
ê half life ç CîÅ is 5750 years, when were ê Dead
Scrolls made?
A)è390 A.D.è B)è130 A.D.è C)è130 B.C.èD)è390 B.C.
üèè Any livïg organism contaïs carbon which is present
ï 2 isoëpes.èOne is ê domïant form, stable Carbon-12
while a small percentage is radioactive Carbon-14.èWhile
ê organism lives, ê percentage ç CîÅ ë Cîì remaïs
constant at ê rate present ï ê air that ê organism
uses ë survive.èWhen ê organism dies, ê radioactive
CîÅ is no longer replenished å due ë its radioactive
decay, its percentage becomes smaller å smaller.èThis
percentage can be measured usïg a Geiger counter.èUsïg
ê half life ç 5750 years, W. F. Libby ï 1952 developed
ê technique ç CARBON DATING ë fïd ê date ç death
ç a livïg object.
èèFirst ê decay rate constant must be found which can
be done by manipulatïg ê half life formula
èèè ln[2]
Tè=è───────
èèèè s
So
èèè ln[2]èèèè ln[2]
sè=è───────è=è───────────è=è0.000121 yearúî
èèèè Tèèèè 5740 years
èèAs 22.3% ç ê Dead Sea Scrolls' CîÅ has been lost, that
means that 77.7% ç ê CîÅ is still present.èSubstitutïg
ïë ê decay function yields'
77.3è=è100eúò°òòòîìî▐
0.773è=èeúò°òòòîìî▐
Takïg ê natural log ç both sides gives
ln[0.773]è=èln[eúò°òòòîìî▐]è=è- 0.000121t
Thus èè tè=è- ln[0.773] / 0.000121
èèèè=è2130 years
èèèè=è 130 B.C.
ÇèC
è5è Six hours after a radioactive sample is produced, 80 grams
ç ê sample are still left.èTwo hours later (8 hours after
production) 50 grams are left.èHow much ç ê sample is
produced?
A)è 500 gramsè B)è328 gramsèC)è299 gramsèD)è167 grams
ü.èèIn this problem, neiêr ê decay rate constant (or ê
equivalant half life) nor ê ïitial amount is given so
êre are two quantities missïg from ê decay function.
è There is ïformation from two times which allows for
developïg 2 equations ï 2 unknowns.èUnfortunately, êy
are not lïear but êy can still be solved.èSubstitutïg
ïë ê decay functions yields ê 2 equations
80è=èP╠ eúÖ6
å
50è=èP╠ eúÖ8
Dividïg ê ëp equation by ê botëm allows P╠ ë be
cancelled leavïg one exponential equation ï one unknown
80èèèP╠ eúæÖèèè eúæÖ
────è= ─────────è=è──────
50èèèP╠ eúôÖèèè eúôÖ
Simplifyïg
1.6è=èeìÖ
Takïg ê natural log ç both sides
ln[1.6] = ln[eìÖ]è=è2s
Thusè s =èln[1.6] / 2 hours =è0.235 hourúî
This can be substituted back ïë eiêr ç ê decay
functions above.èUsïg ê 6 hour value ç 80 grams
80è=èP╠eúò°ìÄÉÑæªè=èP╠eúî°Åîò
Solvïg for P╠ gives
P╠è=è80eî°Åîòè=è328 grams
ÇèB
è6è On Wheel ç Fortune, a person won an annuity valued at
$25,000 ï 10 years.èIf ïterest is compounded contïuously
at 8%, what is ê cost ç ê annuity now?
èèA)è$18,736.40è B)è $15,983.06èC)è$13,131,31èD)è$11,233.22
ü èèThis is a present value problem with s = 0.08,
P = $25,000 å t = 10 years.èSubstitutïg this ï ê
present value equation gives
P╠ = 25000eúò°òôÑîòª
è = 25000eúò°ô
è = $11,233.22
ÇèD
7èSallie McMullï has had a great summer.èFirst she had a
lovely baby boy å ên won $50,000 ï ê lottery.èShe
wants ë ensure that ê baby will have $50,000 at age 18
ë start college.èHow much money must she ïvest at 5%
compounded contïuously ë have $50,000 ï 18 years?
èè A)è $15,942.39è B) $20,328.48èC) $26,008.43èD) $31,345.71
üèèèèThis is a present value problem with s = 0.05,
P = $50,000 å t = 18 years.èSubstitutïg this ï ê
present value equation gives
P╠ = 50000eúò°òÉÑîôª
è = 50000eúò°ö
è = $20,328.48
Thus Sallie will have nearly $30,000 ë spend now å will
still be able ë ensure her son's college education.
ÇèB
è8èPetersburg, Virgïia had a population ç 50,000 ï 1970
å 45,000 ï 1990.èAssumïg an exponential decay model,
estimate ê population ï 2010.
A)è37,942èèB)è39,783èèC)è40,500è D)è41,239
üèè To use ê decay formula with t = 2010 - 1970 = 20 years,
ê ïitial population ç 50,000 is available, but ê decay
rate is not.
è The decay rate can be determïed by usïg ê data from
1990 where t = 1990 - 1970 = 40 years, P╠ = 50,000 å
P = 45,000 å s is ê only variable ï ê function
45000è=è50000eúÖìò
45,000 / 50000è=èeúìòÖ
Takïg ê natural log ç both sides gives
èèèèln[0.9] = ln[eúìòÖ]è=è-20s
So s = - ln[0.90] /20 year =è0.00527 yearúî
èèSubstitute now t = 40 years
Pè=è50000eúÑò°òòÉìƪÅò
è =è50000eúò°ìîî
è =è40,500
Ç C