2.4 just like your checkbook-it all has to balance

8/12/2019 2.4 Just Like Your Checkbook-It All Has to Balance

1/30

A Look at NuclearScience and Technology

Larry Foulke

Atomic and Nuclear Physics The Einstein Connection

2.4 Just like your checkbook; it all has to balance andUnstable nuclides eventually go away


2/30

Nuclear Decay Chart of Nuclides

2

Check the web athttp://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.htmlFor an alternative chart

Try also

http://wwwndc.jaea.go.jp/CN10/

Or

http://en.wikipedia.org/wiki/Chart_of_nuclides

These web links are on the Week 2 Overview.


3/30

Nuclear Data

NZ


4/30


5/30

Nuclear Decay Balance Eqns. Shorthand notation for writing nuclear decay events

Similar to chemical balance equations

Equation must always conserve mass and charge Typically dont list energy or momentum in these

balance equations5

92

235

U!

90

231

Th+ 24

"

Examples

!" 00

0

1

239

94

239

93 ++#

$PuNp

Alpha Decay:

Beta Decay:

Positron Emiss.: !" 0

0

0

1

11

5

11

6 ++#

+

BC


6/30

Nuclear DataN

Z


7/30

Nuclear Decay Balance Eqns. Shorthand notation for writing nuclear decay events

Similar to chemical balance equations


balance equations7

!

4

2

231

90

235

92 +"

ThU

Examples

HeThU

4

2

231

90

235

92 +!

or

93

239Np!

94

239Pu +

"1

0#+0

0$

Alpha Decay:

Beta Decay:


0

0

1

11

5

11

6 ++#

+

BC


8/30

NZ


9/30

Nuclear DataN

Z


10/30

Nuclear Decay

The decay of an unstable nucleus is a random process. Every unstable nuclide is characterized by a unique

decay constant,!

. Decay Constant

The probability that a single nucleus will decay perunit time.

Units:10

!"

#$%

&

secondnuclei

decay


11/30

Decay Activity

IfNis the number of nuclei present in a sample thenthe rate of nuclear decays for the sample is given by:

Ais referred to as the activityof the sample.Activity has basic units of

SI Unit: 1 Becquerel [Bq] = 1 decay/sec Old Unit: 1 Curie [Ci] = 3.7!1010decay/sec

11

[ ]nucleisecondnuclei

decay

second

decay!"

#

$%&

'="#

$%&

'

A

(t)= !N

(t)


12/30

Decay Activity (Time Dependent)

Each radioactive decay destroys one of the unstablenuclei, changing the number of nuclei present.

The number of nuclei,N, and activity,A, are timedependent quantities

12

1decay[ ] = !1nucleus[ ]

( ) ( )tNtA !=


13/30


The original formula can be rewritten in terms of thefractional nuclide population remaining after time t.

13

N t( )=N 0( )e!" t

A t

( )= A 0

( )e!" t

N t( )N 0( )

=A(t)

A(0)= e!" tFractional population

at time t


14/30

Example calculation Presume that we have 1,000,000 nuclei of uranium-235 which

has a half-life of

And lets say we want to know the fractional population ofuranium-235 after radioactive decay for one million years

First, we have to calculate the decay constant from the half-life of7.04x108years

14

!1

2

= 7.04x108years = 704million years

!= 0.693

7.04x108yrs( )

= 9.84x10"10yrs"1


15/30

Example calculation

Presume that we have 1,000,000 nuclei of uranium-235 whichhas a radioactive decay constant of:

And lets say we want to know the fractional population ofuranium-235 after radioactive decay for one-year (or one yearsworth of seconds = 3.15576x107seconds)

The answer is:

15

)( )

te

N

tN !"=

0

Fractional population

at time t

! = 9.84x10"10

yrs"1

N t( )

N 0( )= e

!9.84x10!10yr!1( )(1,000,000yrs)= e!9.84x10

!10

= 0.99902


16/30

Decay in Units of Half-Life

Knief, Fig 2-2


17/30

Half-Life Examples

Uranium 232 70 yr233 160,000 yr234 250,000 yr

235 704,000,000 yr236 23,000,000 yr238 4,500,000,000 yr

Fission ProductsStrontium-90 29 yrCesium-137 30 yr

17


18/30

Mean Time to Decay

In addition to the half-life it is also usefulto know the mean lifetime for a nuclide

Mean Lifetime (")The average (mean) time that it will take for

a single nuclide to decay.

Units: [seconds]18

!"

1=


19/30

Nuclear Decay During a nuclear decay much of the excess energy of an

unstable nuclei is removed with the emitted particle:

Changes in binding energy of nucleus Kinetic energy given to emitted particle

However, following the decay event, the productnucleus may be left in an excited state (still too muchenergy)

In these cases the nucleus can do one of two things: Undergo nuclear decay again Rearrange nucleons in nucleus to achieve a lower overall

energy state. 19


20/30

1. Reprinted with permission from Bechtel MarinePropulsion Corporation.

Image Source Notes


21/30

Supplemental Exercise Slides

21


22/30

Nuclear Decay Balance Eqns.

Shorthand notation for writing nuclear decay events Similar to chemical balance equations


balance equations 22

!4

2

231

90

235

92 +" ThU

ExamplesHeThU

4

2

231

90

235

92 +!

or!" 0

0

0

1

239

94

239

93 ++#

$PuNp

Alpha Decay:

Beta Decay:


0

0

1

11

5

11

6 ++#

+

BC


23/30

NZ


24/30

24

NZ


25/30

Example calculation #2

Presume that we have 1,000,000 nuclei of Tritium (H-3) whichhas a radioactive decay constant of:

And lets say we want to know how long it will take for thefractional population of Tritium to decay to half of it initial value

We find the answer by solving for time, t. We solve and find that

25

( )( )

te

N

tN !"=

0

Fractional population

at time t

! = 1.78x10"9sec

"1

0.5=N t( )N 0( )

= e! 1.78x10

!9sec

!1( ) t( )


26/30

Explanation of natural log

calculations

26

x = e

!yStandard equation used

in this course

We take the natural

log of both sides to get !n x( ) = !n e!y( ) =!y

Which usually looks like thisin our problems:

( )( )

t

eN

tN !"=

0

!n

N t( )N 0( )

!

"#

$

%& = !n e

'( t( ) ='(tWhich usually looks like this

in our problems:


27/30

Decay Half-Life Calculate the amount of time required for 50% of a nuclide

population to decay:

This is referred to as the nuclides half-life Conversely, the decay constant is found by

27

N t( )N 0( )

=

1

2= 0.5

0.5 = e!"T1

2 !n 0.5[ ]=!"T12 T12 = !!n 0.5[ ]

"

T12 =!!n 0.5[ ]

"Units: [seconds]

! ="!n 0.5[ ]

T12

=

0.693

T12


28/30


Looking at units we see:

The decay activity is the rate of decrease of a given nuclidepopulation

Can be written as a first-order separable ODE (OrdinaryDifferential Equation)

28

A t( ) =decays

second

!

"#

$

%& =

-nuclei

second

!

"#

$

%& =

-dN t( )

dt

( ) ( )

( )tNdt

tdNtA !=

"

=


29/30


From this ODE we can solve for the nuclidepopulation as a function of time,N(t)

29

( )( )tN

dt

tdN!=

"

( )( )

dttN

tdN!"=

( )

( ) !! "= dttdNtN

#1

( )[ ] CttNLog +!= "

( ) Ct eetN !"=

Original Equation

Multiply both sides by dt/N(t)

Integrate

Take exponent of both sides

Evaluate indefinite integral


30/30


We need a boundary condition to determine theconstant of integration, C

LetN(0) be the initial nuclide population (at t=0)

Final equation:

( ) Ct eetN !"=

( ) CeeN 00 !"=

( )0NeC =

1

( ) ( ) teNtN !"= 0

Memorize this!

2.4 just like your checkbook-it all has to balance

Documents