Nuclear Physics
The nucleus
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© The scientific sentence. 2010

αdecay and dose absorbed
We are interested in evaluating the absorbed dose
of the αdecays projectile particles emitted from
a Plutonium238, by a sample of water.
1. Plutonium element
Let's consider the Plutonium element:
Symbol: Pu
Atomic Number: 94
Number of Protons/Electrons: 94
Number of Neutrons: 150
Atomic Mass: (244.0) amu
Classification: Rare Earth (actinide series: synthetic or manmade).
Density at 293 K (20^{0}): 19.84 g/cm3 =
19.84 x 10^{3} kg/10^{6} m^{3} =
19.84 x 10^{3} kg/m^{3}
Isoptopes:
 Plutonium244, the most stable, of halflife
of about 80 million years.
 Plutonium240, with a high rate of spontaneous fission.
 Plutonium239, the most useful for nuclear weapons,
fissile as Plutonium241 .
 Plutonium238 has a halflife of 88 years and emits
alpha particles. It is e heat source in thermoelectric generators.
We will focus on the istope 238.
2. Moles and Avogadro number
Avogadro number = 6.023 x 10^{23}
1 uma = 12 x 10^{ 3} /12 N_{av} =
1.66054 x1027 kg.
We neglect the mass of the electrons, thus
the mass considered is the mass of the
nucleus.
1 mol of Pu weights 244.0 grams. That is, 1 gram of Pu
contains 1 mol/238.0 nucleus = N_{av}/238.0 nucleus.
If we consider a mass m (grams) of Pu, then, this mass
corresponds to:
m x N_{av}/238.0 nucleus.
generally,
The number of nuleus N in a mass of m (grams)
of an element of atomic mass "A"(grams) is:
N = m x N_{av}/A nucleus
m and A in grams.
3. Density
The density d is equal to m/V.
For a sphere of diameter D, V = (4/3)*π(D/2)^{3}, we
have:
m = d V = d (4/3)*π(D/2)^{3}
And:
N = d (4/3)*π(D/2)^{3} x N_{av}/A
A 1 µm diameter sphere of Pu contains:
N = d (4/3)*π(D/2)^{3} x N_{av}/A =
19.84 x 10^{3} kg/m^{3} (4/3)*π(1 x 10^{6}/2)^{3} m^{3}
x 6.023 x 10^{23}/238.0 x ^{ 3} kg
= 2.6 x 10^{10} nucleus.
This ball of Pu is going to decay.
4. αdecays from a Plutonium238
At time t_{0} = 0, the number of nucleus that is available is
N_{0} = 2.6 x 10^{10} nucleus.
By decaying, the remaining number of nucleus at the
time t is N(t) = N_{0} exp { λt}
The half of remaining number of nucleus N_{0} /2
corresponds to the halflife t_{h}:
N_{0}/2 = N_{0} exp { λt_{h}}
Then: 1/2 = exp { λt_{h}} , hence:
 log 2 =  λ t_{h}, thus:
t_{h} = log 2 /λ
For the Plutonium238, we have the halflife t_{h} = 88 years
= 88 x 365 x 24 x 60 x 60 seconds = 2.8 x 10^{9} s
Therefore
λ = log 2/ t_{h} = 2.50 x 10^{ 10} s^{1}
5. Activity of the αdecays from a Plutonium238
The activity is:
A(t) =  dN(t)/dt = λ N_{0} exp { λt} = λ N(t)
The initial activity is A(0) = λ N(0) = λ N_{0}
For the Plutonium238, A(0) = 2.50 x 10^{ 10} x 2.6 x 10^{10} =
6.5 nucleus/s = 6.5 decays/sec = 6.5 Bq
Since:
1 Ci = 3.70 x 10^{10} Bq =
3.70 x 10^{10} decays/s,
then:
A(0) = 6.5 Bq = 6.5 Ci /3.70 x 10^{10} =
1.76 x 10^{ 10} Ci
If each α emitted has an energy of 5 MeV =
5 x 10^{6} x 1.6 x 10^{19} joules =
The activity at t = 1 µs is A(1 µs) =
λ N_{0} exp { λ1 µs}
For the Plutonium238, A(1 µs) =
2.50 x 10^{ 10} x 2.6 x 10^{10} exp { 2.50 x 10^{ 10}1 x 10^{ 6}} =
6.5 exp { 2.50 x 10^{ 16}} = 6.5 x 1 = 6.5 decays/sec = 6.5 Bq
= 6.5 Bq = 1.76 x 10^{ 10} Ci
6. Dose absorbed and RBE
If each α emitted has an energy of E = 5 MeV =
5 x 10^{6} x 1.6 x 10^{19} joules =
8 x 10^{13} joules, the dose deposited in a
sphere of 5 µm of diameter in water is:
D = E/V
V = (4/3)π (D/2)^{3} = (4/3)π (5 x 10^{6}/2)^{3}
= 6.55 x 10^{17} m^{3}
The density of water is equal to 1 kg/dm^{3} =
1 kg/ ^{3}m^{3} =
1000 kg/ m^{3} , d = m/V = 1000 kg/ m^{3}
Therefore:
V (in m^{3}) corresponds to a mass of M = 1000 x V Kilograms =
1000 x 6.55 x 10^{17} = 6.55 x 10^{14} Kilograms
Therefore:
D = E/M = 8 x 10^{13} / 6.55 x 10^{14} Joules/Kilograms
= 12.20 Joules/Kilograms
Since:
1 Joule/Kg = 1 Gy
The dose deposited by one α particle of
5 MeV energy due to the decay of a nucleus of Plutonium238,
in a sphere target of water of 5 µm of diameter is 12.20 Gy.
For this nucleus, we have an activity of A(0) = 6.5 αdecays/second
Then: 6.5 α deposit 6.5 x 12.20 Gy = 79.3 Gy/second
1 year = 365 x 24 x 60 x 60 seconds = 3.15 x 10^{7} s, so:
6.5 α deposit 79.3 Gy/second = 3.15 x 10^{7} x 79.3 Gy/year =
2.50 x 10^{9} Gy/year
Since 1 rad = 0.01 J/kg = 0.01 Gy, we have:
D = 12.20 Joules/Kilograms = 12.20 Gy = 12.20 x 100 rad = 1220.0 rad
All the dose deposited is absorbed., therefore,
the Equivalent dose (rem) = RBE x absorbed dose(rad)
The RBE of α particle = 20 rem/rad
Therefore,
The Equivalent dose (rem) of the one α particle emitted
and absorbed by the sample of water is:
20 x 1220.0 rem = 24400 rem
During a year from the Plutonium sphere, we have
2.50 x 10^{9} Gy/year, that corresponds
to the dose:
Dose = 2.50 x 10^{9} Gy/year = 2.50 x 10^{11} rad /year =
2.50 x 10^{11} x 20 rem/year = 5.0 x 10^{12} rem/year
Terminology:
RBE: Relative Biological Effectiveness
in Sv/Gy = rem/rad
Sv = Sievert
rem = Röntgen equivalent for man
1 Gy (gray) = 1 J/Kg
1 rad = 0.01 Gy

