Radioactivity

Lecture 5 The Nature and Laws of Radioactivity

Changing Z to N or N to Z Adding a proton (electron)

Carbon to Nitrogen

Gold to Mercury

Subtracting or adding neutrons

nucleus becomes unstable and decays by internally converting neutrons to protons (beta-decay)!

What are the physical laws that govern the decay process?

198Au

198Hg

14C

14N

Terminology of nuclear decay

Activity: number of decay events per time Decay constant: probability of decay Half life: time for the activity to be reduced to 50%

Time dependent change from configuration 1 (radioactive nucleus) To configuration 2 (decay product, daughter)

Activity corresponds to the number of sand particles dripping through hole

Decay constant is associated with the size of the hole

Units for the Activity A for a radioactive nucleus

1 Ci = 3.7·1010 decays/s = 3.7·1010 Bq

1 Bq = 1 decay/s

Classical unit: Curie: Ci corresponds to the number of decays of 1 g Radium as introduced by Madame Curie

Modern unit: Becquerel: Bq notes a single decay event

Example: the human body is radioactive with an activity of: 2.2⋅10-7Ci = 0.22µCi ⇒ 8000 Bq = 8 kBq Sounds comfortably low sounds alarmingly high

Radioactive Decay Law

( )tdaughter

tmother

eAtA

eAtA

⋅−

⋅−

−⋅=

⋅=

λ

λ

1)(

)(

0

0

λ≡decay constant; a natural constant for each radioactive element. Half life: t1/2 = ln2/λ

exponential decay with time! At half life 50% of the activity is gone!

0.0100.0200.0300.0400.0500.0600.0700.0800.0900.0

1000.0

0 200 400 600 800 1000 1200

time [years]

abu

nd

ance

/act

ivity t1/2=100 years

Describes the change of activity with time

1st example: 22Na 22Na is a radioactive nucleus with a half-life of 2.6 years, what is the decay constant? Mass number A=22; (don’t confuse with activity A(t)!)

197

77

1

2/1

105.81014.36.2

2ln

101014.31

:27.06.22ln2ln

−−

−

⋅=⋅⋅

=

⋅≈⋅=

===

ss

ssy

yyt

λ

π

λ

Radioactive Decay Laws Activity of radioactive substance A(t) is at any time t proportional to number of radioactive particles N(t) :

A(t) = λ·N(t)

A 22Na source has an activity of 1 µCi = 10-6 Ci, how many 22Na nuclei are contained in the source?

(1 Ci = 3.7·1010 decays/s)

1219

1106

19

6

1036.4105.8

107.310105.8

10⋅=

⋅⋅⋅

=⋅

== −−

−−

−−

−

ss

sCiAN

λ

How many grams of 22Na are in the source?

An amount of A grams of atoms with the mass number A (≅1mole) contains NA nuclei

NA ≡ Avogadro’s Number = 6.023·1023 nuclei/mole

➱ 22g of 22Na contains 6.023·1023 nuclei

( )

( ) ggNaN

particlesg

particlesNaN

1023

1222

23

1222

1059.110023.6

1036.42222

10023.61

1036.4

−⋅=⋅

⋅⋅≡

⋅=

⋅=

teNtN ⋅−⋅= λ0)(

How many particles are in the source after 1 y, 2 y, 10 y?

CisyAeyN

CisyAeyN

CisyAeyN

tNstNtAetN

yy

yy

yy

ty

µ

µ

µ

λ

067.05.2490)10(1093.21036.4)10(

58.021590)2(1054.21036.4)2(

765.028305)1(1033.31036.4)1(

)(105.8)()(1036.4)(

1111027.012

112227.012

112127.012

1927.012

1

1

1

1

==⋅=⋅⋅=

==⋅=⋅⋅=

==⋅=⋅⋅=

⋅⋅=⋅=⋅⋅=

−⋅−

−⋅−

−⋅−

−−⋅−

−

−

−

−

Decay in particle number and corresponding activity!

2nd example: Radioactive Decay Plutonium 239Pu, has a half life of 24,360 years. 1. What is the decay constant? 2. How much of 1kg 239Pu is left after t=100, 1,000, 10,000, 24,360, 100,000years?

kgyN

kgyN

kgyN

kgyN

kgyN

ekgyNeNtN

yyt

Pu

Pu

Pu

Pu

Pu

yyPu

tPu

0578.0)000,100(

5.0)360,24(

7520.0)000,10(

9719.0)000,1(

9972.0)100(

1)100()(

1085.224360

2ln2ln

239

239

239

239

239

15

2392391001085.2

0

15

2/1

=

=

=

=

=

⋅=⇒⋅=

⋅===

⋅⋅−⋅−

−−

−−λ

λ

From parent to daughter nuclei

N1 (parent)

N2 (daughter)

)1(

,

02

01

102210

t

t

eNNeNN

NNNNNN

⋅−

⋅−

−⋅=

⋅=

−=+=

λ

λ

N0

The initial radioactive nuclei slowly decay with time converting the initial radioactive species to non radioactive material (or to yet another radioactive daughter nucleus).

14C ⇒ 14N T1/2=5,730 y

22Na ⇒ 22Ne T1/2=2.6 y

26Al ⇒ 26Mg T1/2=716,000 y

40K⇒40Ar T1/2=1,280,000,000 y

3rd example: determine the number of daughter nuclei

Assume a mix of 100 nuclei of 14C, 22Na, 26Al, and 40K each. Calculate the number of daughter nuclei after: t1=10 y, t2=10,000 y, t3=10,000,000 y and t4=10,000,000,000 y

)1()1( 2/1

2ln

002

tTt eNeNN

⋅−⋅− −⋅=−⋅= λ

t 10y 10000 y 10000000 y 10000000000 y T1/2 λ

14C 5730 1.21E-04 1.21E-01 7.02E+01 1.00E+02 1.00E+02 14N 22Na 2.6 2.67E-01 9.30E+01 1.00E+02 1.00E+02 1.00E+02 22Ne 26Al 716000 9.68E-07 9.68E-04 9.63E-01 1.00E+02 1.00E+02 26Mg 40K 1280000000 5.42E-10 5.42E-07 5.42E-04 5.40E-01 9.95E+01 40Ca/40Ar