© JP 1 RADIOACTIVE DECAY 2 It is impossible to say when a particular nucleus will decay. It is only...
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Transcript of © JP 1 RADIOACTIVE DECAY 2 It is impossible to say when a particular nucleus will decay. It is only...
© JP1
RADIOACTIVE RADIOACTIVE DECAYDECAY
© JP2
• It is impossible to say when a particular nucleus will decay .
• It is only possible to predict what fraction of the radioactive nuclei will decay in a certain time.
RADIOACTIVE DECAY OF ATOMS (TRANSMUTATION OF ATOMS) IS A RANDOM PROCESS
RADIOACTIVE DECAY OF ATOMS
(TRANSMUTATION OF ATOMS)
IS A RANDOM PROCESSRANDOMRANDOM
© JP3
Radioactive Decay is not affected by:
1. Physical Conditions - like temperature or pressure.
2. Chemical Changes – it does not matter if the radioactive isotope is part of a
compound.
The half-life of a radioactive isotope is the average time it takes for half of its
atoms to decay.
© JP4
Number of atoms, N
time, t
Initial numb of atoms, N0
20N
HALF LIFE t1/2
t1/2
40N
t1/2
80N
t1/2
THE RATE OF DECAY DEPENDS UPON THE
AMOUNT OF MATERIAL REMAINING, SO THE
DECAY IS EXPONENTIAL
© JP5
N0
N
t
teNN 0
λ = the decay constant – measures the probability of an atom decaying
λ has units of seconds-1
When N = ½ No, t = t1/2 2
1
002
1 t
eNN
22/1 te
2ln
2/1 t
© JP6
A0
A
t
teAA 0
As the Activity, A, depends upon the number of atoms in the source, the Activity also decays exponentially
N.B. Active isotopes have short half lives
© JP7
teNN 0
differentiating teNdt
dN 0
Ndt
dN
orN
t
Nactivity
i.e. The activity = the number of atoms in
the source times the decay constant
© JP8
UNITS OF ACTIVITY
1 Bq = one disintegration per second
Gray (Gy) – the amount of radiation causing 1 kg of tissue to absorb 1 joule (J) of energy
Other units
Sievert (Sv) – Arbitrary unit, based on the Gray , but adjusted to account for damage to living tissue.
© JP9
Example 1
A radioactive source contains 1 x 10-6 g of plutonium – 239. The source is found to emit 2300 alpha particles per second in all directions. Find the half life of plutonium.
1. Finding the number of atoms present in 1 x 10-6 g of plutonium – 239
2. 239 g contain 6.02 x 1023 atoms
3. Hence 1 x 10-6 g contain 2.52 x 1015 atoms = NN
t
N
4. 2300 = λ x 2.52 x 1015
5. λ = 9.13 x 10-13 s-1
132
1 1013.9
693.02ln
t
6. t1/2 = 7.59 x 1011 s = 24 060 years
© JP10
Example 2A compartment on a Geiger Müller tube is filled with a solution containing 1.00g of carbon extracted from one of the Dead Sea scrolls. This gives a count rate of 1000 per hour. When a similar solution containing 1.00 g of carbon extracted from a living plant is used instead, the count rate is 1200 per hour. With no solution in the compartment, the count rate is 300 per hour.
Estimate the age of the scroll if the half life of carbon -14 is 5600 years.
The original count rate corrected for background radiation is A0 = 1200 – 300 = 900 counts per hour
The corrected count rate after the sample has decayed for t years is A = 1000 – 300 = 700 counts per hour
t = 2027 years
teAA 014
2
1
1024.15600
693.02ln yeart
900
70041024.1 te
© JP11
MEASUREMENT OF HALF LIFE
1. SIMPLY MEASURE AND PLOT HOW THE ACTIVITY VARIES WITH TIME
MEASUREMENT OF SHORT HALF LIVES
2. READ OFF
20N
t1/2
REPEAT AND AVERAGE !! OR PLOT A LOG GRAPH
© JP12
MEASUREMENT OF HALF LIFEMEASUREMENT OF SHORT HALF LIVES
LOG GRAPHS
tAA 0lnlnmxcy
teAA 0
lnA
t
gradient = - λ
2ln
2/1 t
© JP13
MEASUREMENT OF HALF LIFE
MEASUREMENT OF LONG HALF LIVES
1. RECORD ACTIVITY
2. BY WEIGHING / CHEMICAL ANALYSIS FIND THE NUMBER OF MOLES OF MATERIAL PRESENT
t
N
3. USING AVOGADRO’S NUMBER, FIND THE NUMBER OF ATOMS PRESENT
4. APPLY TO FIND λNt
Nactivity
2ln
2/1 t