400
7 Nuclear Physics – I
Therefore, Mean free path λ = 1/Σ = 0.2 cm
Fraction of scattered particles = Probability of scattering through more than
90
0 is t/λ = 10
−5
/0.2 = 5 × 10
−5 .
7.19 r min =
R0
2
1 +
+
4b
2
R
2
0
1
2
R 0 = 1.44z Z/T = 1.44 × 1 × 50/(1 × 10
−3 ) = 7.2 × 10
4 fm
bp = L =
b = / p = c/cp = 197.3 MeV fm/1 × 10
−3 MeV = 1.973 × 10
5 fm
b/R 0 = 1.973 × 10
5
/(7.2 × 10
4 ) = 2.74
r min =
7.2
2
× 10
4
1 +
1 + 4 × 2.74
2
1
2
= 2.365 × 10
5 fm
7.20 The mean free path λ = 1/Σ = 1/N .dσ
(1)
where N is the number of atoms per cm
3
N = N 0 ρ/A = 6.02 × 10
23
× 19.3/197 = 5.9 × 10
22
(2)
d σ = (1.44z Z/4T )
2 d Ω/ sin
4 (θ/2)
(3)
Put z = 1, Z = 79, θ = 60
◦
, dΩ = area/(distance)
2
= 0.5/10
2
= 0.005
Using these values in (3), d σ = 2.588 fm
2
= 2.588 × 10
−26 cm
2
(4)
Using (2) and (4) in (1) we find λ = 655 cm,
The probability of scattering at 60
◦ ,
P = t/λ
(5)
where t is the foil thickness
t = pλ = (655 × 1/5 × 10
6 ) cm = 1.31 μm.
7.21 The counting rate is dependent on the factor
(z Z/T )
2 (N 0 /A)(ρt/ sin
4 (θ/2))
In the problem z, T, (ρ t) and θ are unchanged. Hence, counting rate with
platinum/counting rate with silver
= (Z pt
2
/A pt )/(Z Ag
2
/A Ag ) = 1.52
Substituting the known values: Z Ag = 47, A Ag = 108.87 and A pt = 195, the
above equation can be solved to yield Z pt = 77.55 or 78.
7.22 Let the particle of mass m 1 , charge z, velocity v and kinetic energy T collide
elastically with an electron of mass m 2 = m. The electron velocity before
and after the collision is v 2
∗
= v c in the CMS. The velocity v 2
∗ is combined
vectorially with the CMS velocity v c to yield the LS velocity v 2 at angle ϕ
with the incident direction
7 Nuclear Physics – I
Therefore, Mean free path λ = 1/Σ = 0.2 cm
Fraction of scattered particles = Probability of scattering through more than
90
0 is t/λ = 10
−5
/0.2 = 5 × 10
−5 .
7.19 r min =
R0
2
1 +
+
4b
2
R
2
0
1
2
R 0 = 1.44z Z/T = 1.44 × 1 × 50/(1 × 10
−3 ) = 7.2 × 10
4 fm
bp = L =
b = / p = c/cp = 197.3 MeV fm/1 × 10
−3 MeV = 1.973 × 10
5 fm
b/R 0 = 1.973 × 10
5
/(7.2 × 10
4 ) = 2.74
r min =
7.2
2
× 10
4
1 +
1 + 4 × 2.74
2
1
2
= 2.365 × 10
5 fm
7.20 The mean free path λ = 1/Σ = 1/N .dσ
(1)
where N is the number of atoms per cm
3
N = N 0 ρ/A = 6.02 × 10
23
× 19.3/197 = 5.9 × 10
22
(2)
d σ = (1.44z Z/4T )
2 d Ω/ sin
4 (θ/2)
(3)
Put z = 1, Z = 79, θ = 60
◦
, dΩ = area/(distance)
2
= 0.5/10
2
= 0.005
Using these values in (3), d σ = 2.588 fm
2
= 2.588 × 10
−26 cm
2
(4)
Using (2) and (4) in (1) we find λ = 655 cm,
The probability of scattering at 60
◦ ,
P = t/λ
(5)
where t is the foil thickness
t = pλ = (655 × 1/5 × 10
6 ) cm = 1.31 μm.
7.21 The counting rate is dependent on the factor
(z Z/T )
2 (N 0 /A)(ρt/ sin
4 (θ/2))
In the problem z, T, (ρ t) and θ are unchanged. Hence, counting rate with
platinum/counting rate with silver
= (Z pt
2
/A pt )/(Z Ag
2
/A Ag ) = 1.52
Substituting the known values: Z Ag = 47, A Ag = 108.87 and A pt = 195, the
above equation can be solved to yield Z pt = 77.55 or 78.
7.22 Let the particle of mass m 1 , charge z, velocity v and kinetic energy T collide
elastically with an electron of mass m 2 = m. The electron velocity before
and after the collision is v 2
∗
= v c in the CMS. The velocity v 2
∗ is combined
vectorially with the CMS velocity v c to yield the LS velocity v 2 at angle ϕ
with the incident direction