1000 Solved Problems in Modern Physics

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 botheanddare singly charged with the

same kinetic energy the ratio of the radiation loss for deuteron and electron

will be(me/md)^2 ≈(1/ 3 ,670)^2 = 7. 4 × 10 −^8 or≈ 10 −^7

7.34 Ionization loss of muons in the rock=2MeVg−^1 cm^2
=2MeVg−^1 cm^2 ×ρ=2MeVg−^1 cm^2 × 3 .0gcm−^3
=6MeV/cm.
The depth of the rock which will reduce 60 GeV to zero= 60 × 103 /6cm=
104 cm=100 m.

7.35 Ed=mdvd^2 / 2
Ep=mpvp^2 / 2
Ed/Ep=E/(E/2)= 2 =mdvd^2 /mpvp^2 = 2 vd^2 /vp^2
Therefore,vd=vp
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)
TheE/Mratio for the three given particles is identical becauseEp/Mp=
10 / 1 ,Ed/Md= 20 /2 andEα/Mα= 40 /4. Hence
Rd=(RpMdzp^2 )/(z^2 dMp)=(0. 316 × 2 × 12 )/(1× 12 )= 0 .632 mm
Rα=(RpMαzp^2 )/(zα^2 Mp)=(0. 316 × 4 × 12 )/(1× 22 )= 0 .316 mm

7.37 Apply the Bragg–Kleeman formula

RAl/Rair=(ρair/ρAl)

√

AAl/

√

AAir

Substitute the values:ρair= 1. 226 × 10 −^3 gcm−^3 ,ρAl= 2 .7gcm−^3

AAl=27 andAair= 14. 5 ,we findRAl/RAir= 1 / 1 , 614

7.38 The straggling of charged particles is given by the ratioσR/R, whereRis the
mean range of a beam of particles in a given medium andσRis the standard
deviation of the ranges.
NowR=f(v 0 ,I)/z^2 wherev 0 is the initial velocity of beam of particles,Iis
the ionization of the atoms of the absorber. Further,σR=

√

MF(v 0 ,I)/z^2 .So

σR/R=φ(v 0 ,I)/

√

M