A Look at Nuclear
Science and Technology
Larry Foulke
Atomic and Nuclear Physics – The Einstein Connection
2.6 Calculations
Example Problem (Easy)
Krypton-85 has a half-life of roughly 11 years. It’s decay
constant is equal to:
a.
0.063 yrs-1
b.
7.62 yrs
c.
exp [-11(t)]
d.
11 years
Example Problem (Easy)
Krypton-85 has a half-life of roughly 11 years. It’s decay
constant is equal to:
a.
0.063 yrs-1 *
b.
7.62 yrs
c.
exp [-11(t)]
d.
11 years
0.693
λ=
= 0.063 yr −1
11 yrs
Example Problem (Medium)
What is the time required for U-235 to decay by 1%?
Example Problem (Medium)
What is the time required for U-235 to decay by 1%?
Half − life of U − 235 is 7.04x10 8 years
0.693
−10
−1
Decay constant =
=
9.83x10
yr
7.04x10 8 yrs
Now decay by 1% (not decay to 1%)
n(t)
−(9.83x10 −10 )t
= 0.99 = e
n(0)
ln(0.99) = −0.01005 = −9.83x10 −10 (t) or
t = 1.021x10 7 yrs
Example Problem (Harder)
The age of the earth is estimated to be 4.5 billion years. What was the
enrichment of U-235 in natural uranium when the earth was formed?
Example Problem (Harder)
The age of the earth is estimated to be 4.5 billion years. What was the
enrichment of U-235 in natural uranium when the earth was formed?
Half − life of U − 235 is 7.04x10 8 years
U - 235 decay constant =
0.693
= 9.8438x10 −10 yr −1
8
7.04x10 yrs
Half − life of U − 238 is 4.5x10 9 years
U - 238 decay constant =
0.693
−10
−1
=
1.5400x10
yr
4.5x10 9 yrs
.72
= 60.41
0.011918
−10
9
99.2745
99.2745 = N 238 (0)e−(1.5400 x10 )(4.5 x10 ) = N 238 (0)e−0.693or N 238 (0) =
= 198.52
0.5
N 235 (0)
N 235 (0) + N 238 (0) = 60.41 + 198.52 = 258.93 or = 23.3%
N 235 (0) + N 238 (0)
0.72 = N 235 (0)e
−(9.8438 x10 −10 )(4.5 x10 9 )
= N 235 (0)e−4.4297 or N 235 (0) =
Example Problem
Gamma rays interact with Be-9 to produce “photoneutrons.”
What is the threshold gamma energy to make this reaction
happen? The reaction is:
Use the following mass data:
Example Problem
Gamma rays interact with Be-9 to produce “photoneutrons.”
What is the threshold gamma energy to make this reaction
happen? The reaction is: 9 Be + γ → 8 Be + 1n0
Use the following mass data: 9 Be = 9.012182 amu; γ = 0 mass 8
Be = 8.005305 amu; 1n0 = 1.008665 amu so LHS = 9.012182 amu and RHS = 9.013970 amu
Mass difference = 9.013970 − 9.012182 = 0.001788 amu
Mass increases so that requires energy to be supplied
The amount of energy is (0.001788 amu)(931.5MeV / amu) = 1.666Mev
Example Problem
The energy released by the following reaction is most
nearly:
235 U + 1 n -> 2 1 n + 138 Cs + 96 Rb
92
0
0
55
37
a.
b.
Useful data
Element
235 U
92
c.
Atomic mass (amu)
235.043922
138 Cs
55
137.91101
96 Rb
37
95.93427
1 n
0
1.008665
d.
177 MeV
931.5 MeV
236.05 MeV
200 MeV
Example Problem
The energy released by the following reaction is most
nearly:
235 U + 1 n -> 2 1 n + 138 Cs + 96 Rb
92
0
0
55
37
a.
Useful data
Element
235 U
92
Atomic mass (amu)
235.043922
138 Cs
55
137.91101
96 Rb
37
95.93427
1 n
0
1.008665
b.
c.
d.
177 MeV
931.5 MeV
236.05 MeV
200 MeV
LHS : 235.043922 + 1.008665 = 236.052587
RHS : 2(1.008665) + 137.91101 + 95.93427 = 235.86261
Mass loss = 236.052587 − 235.86261 = 0.189977 amu
Mass is lost so energy is released = (0.189977 amu)(931.5MeV / amu)
= 176.96 MeV
Example Problem
Given that the mass of a proton is 1.007825 amu and the
mass of a neutron is 1.008665 amu, the mass of a
deuterium atom (H-2) is __________.
a. 4.032980 amu
b. 3.025155 amu
c. 2.016490 amu
d. 2.014102 amu
Example Problem
Given that the mass of a proton is 1.007825 amu and the
mass of a neutron is 1.008665 amu, the mass of a
deuterium atom (H-2) is __________.
a.4.032980 amu
b.3.025155 amu
c.2.016490 amu
d.2.014102 amu *
According to mass defect, the sum of the constituents of an
atom is great than the observed mass of the given atom.
Answer (c) is equal to the sum of the constituents, (a) and (b)
are greater than the sum, and (d) is less than the sum;
therefore, (d) is the correct answer.
Example Problem
Sodium-22 has a half-life of 2.604 years. Calculate the
activity of a 10 gram sample of pure Na-22.
a. 7.2840×1022 decays/sec
b. 7.2840×1018 decays/sec
c. 5.0793×1015 decays/sec
d. 2.3085×1015 decays/sec
Example Problem
Sodium-22 has a half-life of 2.604 years. Calculate the
activity of a 10 gram sample of pure Na-22.
a. 7.2840×1022 decays/sec
b. 7.2840×1018 decays/sec
c. 5.0793×1015 decays/sec
d. 2.3085×1015 decays/sec*
N Na22 atoms =
gm
10gm
N av atoms / mole =
x6.023x10 23 atoms / mole
gm / mole
22gm / mole
N Na22 atoms = 2.738x10 23
decay constant = λ =
(
0.693
−9
=
8.45x10
sec
7
(2.604yr ) 3.15x10 sec/ yr
Activity = λ N Na22 = 8.45x10 −9
(
)
)(2.738x10 )= 2.31x10
23
15
dis / sec
Example Problem
The Transportation Security Administration stopped a
suspicious passenger and found some radioactive
material on the individual. A 1000 gm sample is found to
have an activity of 2.3×1012 Becquerels. The sample is
most likely:
a) Sr-90
b) Pu-239
c) U-235
d) Cs-137
Example Problem
11. The Transportation Security Administration stopped a
suspicious passenger and found some radioactive
material on the individual. A 1000 gm sample is found to
have an activity of 2.3×1012 Becquerels. The sample is
most likely:
a)Sr-90
1000gms Pu − 239
(6.023x10 atoms / mole)= 2.52x10 atoms
239 gms
/
mole
b)Pu-239 *
Activity = λ N
c)U-235
2.3x10 Bq = 2.3x10 dis / sec
d)Cs-137
2.3x10 dis / sec = λ (2.52x10 atoms )
23
12
12
12
24
2.3x1012 dis / sec
= 9.13x10 −13 sec −1 =
λ=
24
2.52x10 atoms
0.693
T1/2 =
= 7.59x1011 sec = 24,100 yrs
−13
−1
9.13x10 sec
which is close to the half − life of Pu − 239
24