404
7 Nuclear Physics – I
7.3.3 Ionization, Range and Straggling
7.33 The rate of loss of energy due to radiation is inversely proportional to the
square of particle mass and directly proportional to the square of charge of
the incident particle as well as the atomic number of the target nucleus and
directly proportional to the kinetic energy.
−(dE/dx) rad ∝ z
2 Z
2 T /m
2
As the medium is identical and both e and d are singly charged with the
same kinetic energy the ratio of the radiation loss for deuteron and electron
will be(m e /m d )
2
≈ (1/3, 670)
2
= 7.4 × 10
−8 or ≈ 10
−7
7.34 Ionization loss of muons in the rock = 2 MeV g
−1 cm
2
= 2 MeV g
−1 cm
2
× ρ = 2 MeV g
−1 cm
2
× 3.0 g cm
−3
= 6 MeV/cm.
The depth of the rock which will reduce 60 GeV to zero = 60×10
3
/6 cm =
10
4 cm = 100 m.
7.35 E d = m d v d
2
/2
E p = m p v p
2
/2
E d /E p = E/(E/2) = 2 = m d v d
2
/m p v p
2
= 2v d
2
/v p
2
Therefore, v d = v p
Deuteron and proton having the same initial speed will have their ranges in
the ratio of their masses. Therefore the deuteron has twice the range of proton.
7.36 R = (M/z
2 ) f(E/M)
The E/M ratio for the three given particles is identical because E p /M p =
10/1, E d /M d = 20/2 and E α /M α = 40/4. Hence
R d = (R p M d z p
2 )/(z
2
d M p ) = (0.316 × 2 × 1
2 )/(1 × 1
2 ) = 0.632 mm
R α = (R p M α z p
2 )/(z α
2 M p ) = (0.316 × 4 × 1
2 )/(1 × 2
2 ) = 0.316 mm
7.37 Apply the Bragg–Kleeman formula
R Al /R air = (ρ air /ρ Al )
√
A Al /
√
A Air
Substitute the values: ρ air = 1.226 × 10
−3 g cm
−3
, ρ Al = 2.7 g cm
−3
A Al = 27 and A air = 14.5, we find R Al /R Air = 1/1, 614
7.38 The straggling of charged particles is given by the ratio σ R /R, where R is the
mean range of a beam of particles in a given medium and σ R is the standard
deviation of the ranges.
Now R = f (v 0 , I )/z
2 where v 0 is the initial velocity of beam of particles, I is
the ionization of the atoms of the absorber. Further, σ R =
√
M F(v 0 , I )/z
2 . So
σ R /R = ϕ(v 0 , I )/
√
M
7 Nuclear Physics – I
7.3.3 Ionization, Range and Straggling
7.33 The rate of loss of energy due to radiation is inversely proportional to the
square of particle mass and directly proportional to the square of charge of
the incident particle as well as the atomic number of the target nucleus and
directly proportional to the kinetic energy.
−(dE/dx) rad ∝ z
2 Z
2 T /m
2
As the medium is identical and both e and d are singly charged with the
same kinetic energy the ratio of the radiation loss for deuteron and electron
will be(m e /m d )
2
≈ (1/3, 670)
2
= 7.4 × 10
−8 or ≈ 10
−7
7.34 Ionization loss of muons in the rock = 2 MeV g
−1 cm
2
= 2 MeV g
−1 cm
2
× ρ = 2 MeV g
−1 cm
2
× 3.0 g cm
−3
= 6 MeV/cm.
The depth of the rock which will reduce 60 GeV to zero = 60×10
3
/6 cm =
10
4 cm = 100 m.
7.35 E d = m d v d
2
/2
E p = m p v p
2
/2
E d /E p = E/(E/2) = 2 = m d v d
2
/m p v p
2
= 2v d
2
/v p
2
Therefore, v d = v p
Deuteron and proton having the same initial speed will have their ranges in
the ratio of their masses. Therefore the deuteron has twice the range of proton.
7.36 R = (M/z
2 ) f(E/M)
The E/M ratio for the three given particles is identical because E p /M p =
10/1, E d /M d = 20/2 and E α /M α = 40/4. Hence
R d = (R p M d z p
2 )/(z
2
d M p ) = (0.316 × 2 × 1
2 )/(1 × 1
2 ) = 0.632 mm
R α = (R p M α z p
2 )/(z α
2 M p ) = (0.316 × 4 × 1
2 )/(1 × 2
2 ) = 0.316 mm
7.37 Apply the Bragg–Kleeman formula
R Al /R air = (ρ air /ρ Al )
√
A Al /
√
A Air
Substitute the values: ρ air = 1.226 × 10
−3 g cm
−3
, ρ Al = 2.7 g cm
−3
A Al = 27 and A air = 14.5, we find R Al /R Air = 1/1, 614
7.38 The straggling of charged particles is given by the ratio σ R /R, where R is the
mean range of a beam of particles in a given medium and σ R is the standard
deviation of the ranges.
Now R = f (v 0 , I )/z
2 where v 0 is the initial velocity of beam of particles, I is
the ionization of the atoms of the absorber. Further, σ R =
√
M F(v 0 , I )/z
2 . So
σ R /R = ϕ(v 0 , I )/
√
M