468
8 Nuclear Physics – II
When the target and the projectile are interchanged, the same excitation
energy W
∗ produced with T a is given by
−W
∗
= E Ta
m p
m p + m Ta
(2)
Comparing (1) and (2)
E Ta + E p
m Ta
m p
= 897.7 MeV
8.64 E th = |Q|
1 +
m o
m ca
= 7.83
1 +
16
48
= 10.44 MeV
The velocity, β =
υ
c
=
2T
mc 2 =
2 × 10.44
16 × 931
= 3.74 × 10
−2
The impact parameter, b = R 1 + R 2 = 1.1
16
1 / 3 + 48
1 / 3
= 6.153 fm
J = M o υb = n
n =
M o υb
=
M o c
2
βb
c
=
16 × 931 × 3.74 × 10
−2
× 6.153
197.3
= 17.37 or 17
8.3.11 Cross-sections
8.65 If 1% of neutrons are absorbed then 99% are transmitted. The transmitted
number I are related to the incident number by
I = I 0 exp(−μx)
( 1 )
where μ is the absorption coefficient and x is the thickness of the foil.
μ = Σ = σ N = σ N 0 ρ/A
where σ is the microscopic cross-section, N 0 is the Avagardro’s number, ρ the
density and A the atomic weight.
μ =
28000 × 10
−24
× 6.02 × 10
23
× 7.3
114.7
cm
−1
= 1,073 cm
−1
I /I 0 = 99/100 = exp(−1,073x)
x = 9.37 × 10
−6 cm or 9.37 μm
8.66 Specific activity, that is activity per gram
|dQ/dt| = Q max λ = φ Σ act = φσ a N 0 /A
= 5 × 10
12
× 20 × 10
−24
× 6.02 × 10
23
/60
= 10
12 disintegrations per second per gram.
8.67 |dQ/dt| = Qλ = φΣ act (1 − e
−λt )
Given |dQ/dt|/|dQ s /dt| = 20/100 = 1 − e
−0.693t/5.3
or exp(0.693t/5.3) = 1.25
Take log e to find t = 1.7 years.
8 Nuclear Physics – II
When the target and the projectile are interchanged, the same excitation
energy W
∗ produced with T a is given by
−W
∗
= E Ta
m p
m p + m Ta
(2)
Comparing (1) and (2)
E Ta + E p
m Ta
m p
= 897.7 MeV
8.64 E th = |Q|
1 +
m o
m ca
= 7.83
1 +
16
48
= 10.44 MeV
The velocity, β =
υ
c
=
2T
mc 2 =
2 × 10.44
16 × 931
= 3.74 × 10
−2
The impact parameter, b = R 1 + R 2 = 1.1
16
1 / 3 + 48
1 / 3
= 6.153 fm
J = M o υb = n
n =
M o υb
=
M o c
2
βb
c
=
16 × 931 × 3.74 × 10
−2
× 6.153
197.3
= 17.37 or 17
8.3.11 Cross-sections
8.65 If 1% of neutrons are absorbed then 99% are transmitted. The transmitted
number I are related to the incident number by
I = I 0 exp(−μx)
( 1 )
where μ is the absorption coefficient and x is the thickness of the foil.
μ = Σ = σ N = σ N 0 ρ/A
where σ is the microscopic cross-section, N 0 is the Avagardro’s number, ρ the
density and A the atomic weight.
μ =
28000 × 10
−24
× 6.02 × 10
23
× 7.3
114.7
cm
−1
= 1,073 cm
−1
I /I 0 = 99/100 = exp(−1,073x)
x = 9.37 × 10
−6 cm or 9.37 μm
8.66 Specific activity, that is activity per gram
|dQ/dt| = Q max λ = φ Σ act = φσ a N 0 /A
= 5 × 10
12
× 20 × 10
−24
× 6.02 × 10
23
/60
= 10
12 disintegrations per second per gram.
8.67 |dQ/dt| = Qλ = φΣ act (1 − e
−λt )
Given |dQ/dt|/|dQ s /dt| = 20/100 = 1 − e
−0.693t/5.3
or exp(0.693t/5.3) = 1.25
Take log e to find t = 1.7 years.