7.3 Solutions 405
As the straggling is inversely proportional to the square root of particle
mass, the straggling for^3 He will be greater than that for^4 He of equal range.7.39 dE/dR=k/(βc)^2
dR=[(βc)^2 /k]d(Mv^2 /2)=k′Mf(v)dv
wherek′is another constant
Integrating from zero tov 0R=∫
dR=k′M∫
f(v)dv=k′MF(v 0 )If two single charge particles of massesM 1 andM 2 , be selected so that their
initial velocities are identical then their residual ranges
R 1 /R 2 =M 1 /M 2
M 2 =R 2 M 1 /R 1 = 165 × 1 , 837 me/ 1 , 100 = 275. 5 me
It is a pion. Its accepted mass is 273me7.40 Balancing the centripetal force with the magnetic force at the point of
extraction
mv^2 /r=zevB (1)
which gives us
v=zeBr/m
The ratio of velocities forα-particle and deuteron is
vd/vα=(zd.mα)/(md.zα)=(1×2)/(2×1)= 1
sincemα≈ 2 md
As the initial velocities are identical the ratio
Rd/Rα=(mdzα^2 )/(mαzd^2 )= 22 / 2 = 2
Therefore,Rd= 2 Rα
7.41 Use Geiger’s rule
R=KE^3 /^2 (1)
8. 6 =K(8.8)^3 /^2
K= 0. 33
At a distance of 4 cm from the source, residual range is 4.6 cm. At this point
the energy can be found out by applying (1) again
4. 6 = 0. 33 ×E^31 /^2
OrE 1 = 5 .79 MeV
Differentiating (1)
dR/dE=(3/2)KE^1 /^2
dE 1 /dR= 2 / 3 KE 11 /^2 = 2 /(3× 0. 33 ×√
5 .79)
= 0 .84 MeV/cm