P H P 4 8 0 : Nu c l e a r P h ys i c s
C o u r s e I n c h a r g e : Dr B a s k a r a n R a n g a s a m y
Study Material for Week 4 (13.04.2020 to 17.04.2020)
Problems in Half-Life period
Example 1: The half-value period of radium is 1590 years. In how many years will one gram of pure
element (a) lose one centigram, and (b) be reduced to one centigram?
Solution: Here, half-life period of radium = T1/2 = 1590 years.
.
Radioactive constant = 𝜆 =
/
.
=
(a) Let t be the time in which one gram of radium loses one centigram (0.01g).
Radium left behind = 1 - 0.01 = 0.99 gram.
Now, N= N0 𝑒
𝑜𝑟 𝑙𝑜𝑔 𝑁 = 𝑙𝑜𝑔 𝑁 − 𝜆𝑡
𝜆𝑡 = 𝑙𝑜𝑔 ( )
𝑡=
𝑙𝑜𝑔
=
𝑙𝑜𝑔 (
.
)
.
= 23.25 years.
(b) Here, N= 0.01 gram; N0 = 1 gram; t = ?
𝑡=
𝑙𝑜𝑔
=
𝑙𝑜𝑔
.
= 10560 𝑦𝑒𝑎𝑟𝑠
.
Example 2: 1 gram of radium is reduced by 2.1 mg in 5 years by α -decay. Calculate the half-life period
of radium.
Solution: Initial mass of radium = 1 gram
Mass of radium left behind after 5 years = 1 - 0.0021 = 0.9979 g.
We have, 𝑁 = 𝑁 𝑒
; 𝐻𝑒𝑟𝑒,
0.9979 = 𝑒
5𝜆 = 𝑙𝑜𝑔 (
𝜆=
𝑙𝑜𝑔
𝑜𝑟 𝑒
.
=
=
; 𝑡 = 5 𝑦𝑒𝑎𝑟𝑠
.
)
.
=
.
.
𝑙𝑜𝑔 (
.
)
= 41.45 x 10-5 per year.
𝑇/ =
.
=
.
.
×
= 1672 𝑦𝑒𝑎𝑟𝑠
Example 3: Calculate the time required for 10% of a sample of thorium to disintegrate. Assume the halflife of thorium to be 1.4 x 1010 years.
Solution: We have, N= N0𝑒
.
𝜆=
=
/
.
𝑡=
. ×
𝑎
. ×
0.9𝑁 = 𝑁 𝑒
. Here N= 0.9 N0.
𝑜𝑟 𝜆𝑡 = 𝑙𝑜𝑔
× 𝑙𝑜𝑔 1.111 =
.
𝑜𝑟 𝑡 =
.
. ×
𝑙𝑜𝑔 ( . )
× 2.302 × 𝑙𝑜𝑔 1.111 = 2.1 × 10 𝑦𝑒𝑎𝑟𝑠
.
The Mean Life
It is not possible to predict which atom of a radioactive substance will disintegrate at any instant.
The atom which disintegrates first has zero life and that disintegrates last has infinite life. Thus the life of
every atom is different and the actual lives of the various atoms range from zero to infinity.
𝑇ℎ𝑒 𝑚𝑒𝑎𝑛 − 𝑙𝑖𝑓𝑒 𝑜𝑓 𝑎 𝑟𝑎𝑑𝑖𝑜𝑎𝑐𝑡𝑖𝑣𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 =
Definition. The mean-life of a radioelement is defined as the ratio of the total life time of all the
radioactive atoms to the total number of such atoms in it.
Value of mean-life. Let N, be the total number of radioactive atoms in the beginning. Let N be the
number of atoms of that element after time t. Then 𝑁 = 𝑁 𝑒 . Let dN be the number of atoms
disintegrating between time t and t + δt. These dN atoms have had a life between t and (t + δt). Since δt is
very small, each of these atoms had a life of t.
total life of dN atoms = (dN)t.
The possible life of any of the total number N0, radioactive atoms varies from 0 to ∞.
Total life time of all N0 atoms = ∫ 𝑡 𝑑𝑁
Now, mean life = 𝑇 =
=
Now
∫
𝑁= 𝑁𝑒
= −𝜆𝑁 𝑒
𝑑𝑁 = −𝜆𝑁 𝑒
𝑑𝑁 = 𝜆𝑁 𝑒
𝑑𝑡
𝑑𝑡
(Leaving the negative sign which merely indicates the decrease in the number of atoms with time)
Hence,
𝑇=
∫
Integrating by parts, 𝑇 = 𝜆[
= 𝜆 ∫ 𝑡𝑒
− ∫
]
𝑑𝑡
= 𝜆[
−
] = 𝜆
=
𝑇=
Thus the mean life (𝑇) of a radioactive substance is the reciprocal of the decay constant (𝜆).
Example 1: 1 gram of a radioactive substance disintegrates at the rate of 3.7 X 1010 disintegrations per
second. The atomic weight of the substance is 226. Calculate its mean life.
Solution: Number of atoms disintegrated in one second = 3.7 X 1010
The mass of the substance disintegrated in one second =
( . ×
.
)×
×
= 1.389 × 10-14 kg
N = 1 g = 10-3 kg; −
Here,
We have, −
= 1.389 × 10-14 kg
= λN
/
𝜆=
Mean life
𝑇=
=
.
=
.
×
= 1.389 × 10
𝑠
𝑠 = 2282 𝑦𝑒𝑎𝑟𝑠
×
Units of Radioactivity
Generally we express the activity of a radioactive substance in terms of curie and its submultiples
curie (mCi) and micro curie (μCi). The curie is defined as the quantity of a radioactive substance which
gives 3.70 X 1010 disintegrations / second. This is the number of disintegrations per second per gram of
radium. Another unit rutherford (Rd) is also used. It is defined as the quantity of a radioactive substance
which gives 106 disintegrations / second. It is to be noted that for a substance with very short hall-life
period, very little of the substance is required for 1 curie of activity. On the other hand, for the substance
with very long half-life period, a very large quantity of the substance is required for 1 curie of activity.
Example. Calculate the weight in kg of one curie of Ra B (Pb214) from the half-life of 26.8 minutes.
Solution. 1 curie = 3.7 x 1010 disintegrations / sec.
Here,
−
𝑁=
𝑁=
= 𝜆𝑁 = 3.7 X 1010
. ×
=
. ×
(
. ×
×
/
.
. ×
.
)
= 8.585 × 10
𝑎𝑡𝑜𝑚𝑠.
From Avogadro's hypothesis, 6.023 x 1026 atoms are contained in 214 kg of RaB.
8.585 x 1013 atoms are contained in
× .
.
×
×
= 3.1 X 10-11 kg.
𝑘𝑔
Activity. The activity of a sample of any radioactive nuclide is the rate at which the nuclei of its
constituent atoms decay. If N is the number of nuclei present in the sample at a certain time activity R is
given by R=−
. The S.I. unit of activity is named after Henri Becquerel.
1 becquerel = 1 Bq = 1 event/s.
1 MBq = 106 Bq and 1 GBq = 109 Bq.
Example. Find the activity of 1 mg (10-6kg) of radon.
Solution. The decay constant of radon is
𝜆=
.
.
=
.
. ×
= 2.1 × 10 𝑠
The No. of atoms in 10-6 kg of Rn222 = N =
.
×
×
= 2.7 × 10
R = 𝜆𝑁 = (2.1 × 10 ) × (2.7 × 10 )
= 5.7 x 1012 events / s
= 5.7 x 103 GBq = 153 Ci
In this case, the "events” are alpha decays.
Law of Successive Disintegration (Growth of daughter activities)
Chain disintegration means a process in which a radioactive substance disintegrates to form a new
substance, which disintegrates to form another new substance and so on. Consider a substance A which
decays to form a substance B. Let the substance B decay to form a substance C and so on.
𝐴 → 𝐵 → 𝐶 → ⋯ 𝑋 (𝑠𝑡𝑎𝑏𝑙𝑒).
In a radioactive series, any two adjacent elements are considered as parent and daughter. Evidently in a
series, the parent of the following element will be daughter of the preceding one.
At time t = 0, the number of initial atoms in A = N0
and the number of initial atoms in B = 0
At time t, let the number of atoms in A=N1
and the number of atoms in B = N2
Let λ1 and λ2 be the decay constants of A and B respectively. Every time an atom of A disappears,
an atom of B is produced.
Rate of formation of daughter B = λ1 N1
The rate at which В decays = λ2 N2
The net increase in the number of B atoms
= 𝜆 𝑁 −𝜆 𝑁
Hence
= 𝜆 𝑁𝑒
− 𝜆 𝑁 (𝑠𝑖𝑛𝑐𝑒 𝑁 = 𝑁 𝑒
)
+ 𝜆 𝑁 = 𝜆 𝑁𝑒
Multiplying both sides by the integrating factor 𝑒
𝑒
or
+ 𝜆 𝑁𝑒
𝑁𝑒
Integrating,
𝑁𝑒
,
= 𝜆 𝑁 𝑒(
= 𝜆 𝑁 𝑒(
)
𝑁 𝑒(
)
=
)
(1)
+𝐶
The constant of integration C can be determined from the initial conditions.
i.e., when t = 0, N2 = 0; therefore, 𝐶 =
[𝑒 (
)
−𝑒
]
− 1]
Substituting the value of C in equ (1),
𝑁𝑒
=
[𝑒
(2)
Radioactive equilibrium:
Case (i). Secular or permanent equilibrium. Suppose T1 >> T2 with T1≈ ∞ and T2 = 0. (i.e., the half-life
of A is very much longer than that of B).
Then, 𝜆 ≪ 𝜆 and 𝜆 ≈ 0.. In equation (2)
(2), 𝑒
𝑁 =
[1 − 𝑒
→1
(3)
]
After an appreciably long time, 𝑒
becomes negligible and
𝑁 ≈𝑁
𝑁𝜆 =𝑁 𝜆
.
Fig.1: Experimental decay and recovery curves
(4)
This relation shows that at equilibrium, the rate of decay of any radioactive product is just equal to
its rate of production from the previous member of the chain. Now the daughter is said to be in secular or
permanent equilibrium with the parent. An example of this is provided by the decay of radium into radon
according to the equation, Ra226 → Rn222 + α.
The half-life period of Ra (1590 years) is very large compared with that of Rn (3.8 days). Fig. 1
shows the experimental decay and recovery curves. After a time, long enough compared with its mean
life, Rn is in permanent equilibrium with Ra. Its amount becomes constant as shown by the sum of the
ordinates of the two curves which becomes constant beyond a certain value of time indicated by the point
of intersection.
Case (ii). Transient equilibrium. Suppose 𝜆 ≪ 𝜆 but 𝜆 ≠ 0 i.e., the decay constant 𝜆 which is very
small compared with 𝜆 , is not small enough to be ignored. In this case, after a sufficiently long time,
𝑒
becomes negligible compared with 𝑒
𝑁 =
𝑒
. Hence
=
(5)
=
In this case, both A and B decay while the ratio N2/N1 remains constant.
Radioactive Dating: The Age of the Earth
The age of the earth is estimated from the relative abundance of the two isotopes of uranium, U238
and U235. The half-periods of U238 and U235 are 4.5 x 109 years and 7 x 108 years respectively. Assume
that at the beginning when the earth was formed the proportions of the two isotopes were equal. The
present relative abundance of U238 to U235 in natural uranium is 99.3% to 0.7%
=
Where
𝜆 =
.
.
𝑎𝑛𝑑 𝜆 =
. ×
.
×
.
𝑙𝑜𝑔
.
×
)
= (𝜆 − 𝜆 )𝑡
.
𝑡=
= 𝑒(
=
.
𝑙𝑜𝑔
=
.
(
.
. ×
.
)
𝑙𝑜𝑔
.
.
= 5.93 x 109 years.
This value agrees nearly with that given by astronomical evidence for the age of the universe.
Radio-carbon Dating. One of the interesting properties of radio nuclides is that they have constant halflife irrespective of the state of chemical combination of the atoms in which they reside. This property
finds applications in dating or age estimation. 6C14, a carbon isotope is used in archaeological dating. This
isotope has a half-life of 5730 years. The interaction of cosmic rays with the atmospheric gas atoms
produces the radioactive isotope C14. This is assimilated by living organism along with the more abundant
C12. The principle of this method is based upon the fact, that our atmosphere is constantly being
bombarded by cosmic rays which are highly penetrating. These rays mostly consist of high energy
particles like neutrons and protons. Atmospheric nitrogen -14 absorbs fast moving neutrons and changes
into C14 with the emission of a proton;
1
0n
+ 7N14 → 6C14 + 1H1
Carbon-14, which is neutron rich, is radio-active and decays by β-emission to nitrogen-14 with a half life
of 5730 years.
14
6C
→ 7N14 + β- + ν
Thus the carbon dioxide molecules in the air contain a small fraction of radio-active carbon-14
atoms in addition to the stable carbon-12 atoms. Living organisms like plants, trees etc, consume
molecules of carbon dioxide which contain carbon-12 as well as radioactive carbon-14. The production of
14
6C
and its subsequent decay are in equilibrium in the earth's atmosphere. Hence 6C14 is present in
equilibrium concentration in all living plants. As soon as the organism dies, the intake of 6C14 stops and
the radio-active C14 starts decaying according to equation 𝑁 = 𝑁 𝑒
14
with half life of 5730 years.
12
Hence the number of C atoms to the number of stable C atoms in the object decreases with time after
death. Thus the time elapsed since death can then be estimated by measuring this ratio. Using this
procedure, ages of wooden slabs from tombs, charcoal used for old paintings etc., have been studied.
Thus if the relative numbers of the two isotopes of carbon in a given sample are known, the number of
years the organism has been dead, can be determined. The amount of carbon-14 is determined by
measuring its activity i.e., λN = A.
The same technique has been used in estimating the age of the earth from measurement on the
relative amounts of U238 and Pb206 in geological specimens. In 1913 Joly and Rutherford suggested that if
igneous rock, formed as a result of a prehistoric volcanic eruption, contained a small amount of uranium it
would steadily decay, leaving less uranium and depositing more stable Pb-206. By measuring the ratio of
uranium to lead in rock samples, a rather exact time can be determined for the origin of the geological
deposits. Uranium dating measures times of the order of millions of years.
*************