phys 1030 final exam - physics.umanitoba.ca

PHYS 1030 Final Exam

Thursday, April 26

9:00 - 12:00

Frank Kennedy Brown Gym

The whole course, about same weight per chapter

30 multiple-choice questions

Formula sheet provided (also on web site)

Bring student ID!

Check your marks on the web site

Let me know if there are errors!

20Monday, April 9, 2007

Radioactive Decay

Half-life, T1/2

: the time for half of the nuclei to decay.

N0/2

N0/4

N0/8

N0

Start with N0 unstable nuclei

Observe how many survive to time t

After each succeeding half-life,

half of the remaining unstable

nuclei remain...


Radioactive Decay

After time T1/2,

[1

2

]N0 are left

After time 2T1/2,

[1

2

][1

2

]N0 =

[1

2

]2N0 are left

After time 3T1/2,

[1

2

][1

2

]2N0 =

[1

2

]3N0 are left

After time nT1/2,

[1

2

]nN0 are left

= number of half-livesn=t

T1/2So, N(t) = N0

[1

2

]n


Radioactive Decay

= number of half-livesn=t

T1/2So, N(t) = N0

[1

2

]n

This is the same as an exponential decay:

N(t) = N0

[1

2

]n= N0e

−!t ! = “decay constant”

So, !=ln2

T1/2=0.693

T1/2

and, n=t

T1/2Take logs: −n ln2=−!t

Natural log, loge

= number of half lives elapsed

=1

“mean life”!=

0.693

T1/2

t ln2

T1/2= ! t∴


Some Half Lives

Isotope Half LifeDecay

Mode

214Po 0.164 ms ", #

89Kr 3.16 min $–, #

222Rn 3.83 days ", #

60Co 5.271 y $–, #

90Sr 29.1 y $–

226Ra 1600 y ", #

14C 5730 y $–

238U 4.47!109 y ", #

115In 4.41!1014 y!!! $–


Radioactive Decay, Activity

Activity, A = number of decays per second = – %N/%t, N = no. of nuclei left

Statistically, the rate of decay is proportional to the number of radioactive

nuclei present.

Units: Becquerel (Bq): 1 Bq = 1 decay per second

" Curie (Ci, old unit): 1 Ci = 3.7 x 1010 Bq = activity of 1 g of pure

" radium

Short half life means large decay constant and large activity (# = 0.693/T1/2).

Radioactive Decay, Activity

initial activity, A0 = ! N

0

A=−!N!t

= "N(t) = "N0e−"t

A = −dNdt

=− d

dtN0e

−!t

= !N0e−!t

Calculus


Radioactive Decay

!=0.693

T1/2

or

N(t) = N0e−!t, != decay constant

N(t) = N0

[1

2

]nn=

t

T1/2= number of half-lives

The number of radioactive nuclei left after time t is:

Activity, A(t) = !N(t)


A Radioactive

Decay Series

A

Z

238

92U→ 234

90Th+ 4

2He

234

90Th→ 234

91Pa+ e−+ !̄

and so on...

...ending at 20682Pb

All formed in a supernova

explosion about 4.5

billion years ago.31.55

Other decay series:

235U→ 207

Pb

232Th→ 208

Pb

222Rn

238U

Radon, of

basement fame


Radon in the Basement

222Rn – produced in the decay chain that starts with 238U.

222Rn half life is only 3.83 days, but it is generated continually by the

decay of longer-lived nuclei.

N = N0

[1

2

]n, n=

t

T1/2=31

3.83= 8.094

N = N0

[1

2

]8.094= 0.00366 N0 = 1.1×105

The initial activity of the radon is

A0 = !N0 =0.693 N0T1/2

=0.693×3×107

3.83×24×3600 s = 62.8 Bq

31.34

Suppose 3!107 radon nuclei are trapped in a basement when the walls are

sealed so no more can enter. How many are left after 31 days?


Prob. 31.32/34: Strontium 90Sr has a half-life of 29.1 years. It is

chemically similar to calcium, enters the body through the food chain

and collects in the bones. Consequently, 90Sr is a particularly serious

health hazard.

How long will it take for 99.99% of the 90Sr released in a nuclear

reactor accident to disappear?


Radioactive Dating

The best known is radiocarbon dating, based on the decay of 14C.

“Carbon-based life forms” take up carbon in food or as CO2 (plants, trees,

in photosynthesis).

One atom in 8.3!1011 of carbon has a 14C nucleus, the rest are 12C, 13C

(activity 0.23 Bq per gram of C).

14C has a half-life of 5730 years, 12C, 13C are stable.

When the organism dies, the uptake of carbon ceases, and the amount of 14C present decreases, halving every 5730 years.

Measure how much 14C is left & work out how long since organism died.


2003 Final, Q29: The ratio of the abundance of 14C to 12C in a sample

of dead wood is one quarter the ratio for wood in a living tree. If the

half life of 14C is 5730 years, which of the following expressions

determines how many years ago the wood died?

a) 2!5730 b) 4!5730 c) 0.75!5730 d) 0.50!5730 e) 0.25!5730


Prob. 31.41: The practical limit for radiocarbon dating is about 41,000

years. What fraction of the 14C is left after this time? (half-life = 5730

years)

The number of half-lives that have elapsed in 41,000 years is:

n = 41,000/5,730 = 7.16

and so the fraction of 14C left after 41,000 years is

N/N0 = (1/2)7.16 = 0.007

so, only 0.7% of the 14C is left.


Radioactive Dating

The amount of 14C left can be measured by:

• Counting the rate of decay (activity) of 14C – the accuracy of the

" age measurement depends on the size of the sample and on for how

" long you are willing to count. The more decays seen, the more

" accurate the measurement.

• Counting the number of 14C nuclei directly by vaporizing the

" sample and counting the 14C nuclei in a mass spectrometer. You are

" no longer waiting for nuclei to decay and can get much higher

" precision on the age.

31.41


Radioactive dating – origin of the 14C

The 14C comes from cosmic rays that interact with 14N in the upper

atmosphere:

n + 14

7N→ 14

6C + p

Stable, the usual form of N

Number of nucleons: 1 + 14 = 14 + 1

Charge: " 0 + 7 = 6 + 1

The 14C combines with O2 to form 14CO

2 which mixes with normal CO

2

in a stable proportion (1 in 8.3!1011).

14C decays back to 14N:14

6C → 14

7N + e

−+ !̄


Oldest rocks, t = 3.7!109 y; meteorites, moon rocks, t = 4.5!109 y.

+ other methods based on ratios of isotopes.

Radioactive Dating on Geological Timescales

238U & 206Pb in decay series and T1/2

= 4.5 ! 109 years for 238U.

N0 = N(t) +N!

8N(t) = number of U nuclei present now

Then, N(t) = N0

[1

2

]n, n=

t

T1/2→ t

Number of "-particles produced in the decay series is (238 – 206)/4 = 8.

If the decays occur in rock, the 4He can be trapped. Measure how much 4He

is present in the rock. Each U decay should generate a total of 8 "-particles.

Then, the original number of 238U nuclei in the rock when it was formed is:

NowNow


31.40/-: Material found with a mummy in the arid highlands of southern

Peru has a 14C activity per gram of carbon that is 78.5% of the activity

present initially. How long ago did this individual die?

(half life of 14C is 5730 y).


31.52/40: An archeological specimen containing 9.2 g of carbon has

an activity of 1.6 Bq. How old is the specimen?


Nuclei

• Made up of nucleons (neutrons + protons), and held together by the

" strong nuclear force, which is of short range.

• Radius r = r0A1/3, r

0 = 1.2!10-15 m.

• Binding energy, B = energy to separate neutral atom into neutrons and

" hydrogen atoms. B = %m c2, %m = mass defect.

• Unstable nuclei decay to objects of lower total mass, converting the

" difference in mass to energy:

A

ZX→ A−4

Z−2Y+ 42He + energy

A

ZX→ A

Z+1Y+ e−+ !̄+ energy

A

ZX→ A

Z−1Y+ e+ + !+ energy

" #-decay: A

ZX∗ → A

ZX+ !+ energy

' "-decay:

" $-decay:


Radioactive Decay

• Half life, T1/2

= time for half of the radioactive nuclei to decay.

• Number of nuclei left after time t:

N(t) = N0

[1

2

]n= N0e

−!t

n = number of half-lives =t

T1/2

! = decay constant =0.693

T1/2

• Activity = rate of decay, A = –%N/%t

A= ! N(t)

" 1 Bq = 1 decay/second


phys 1030 final exam - physics.umanitoba.ca

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