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 = [(β c)
2
/k] d(Mv
2
/2) = k
M f (v) dv
where k
is another constant
Integrating from zero to v 0
R =
dR = k
M
f (v) dv = k
M F(v 0 )
If two single charge particles of masses M 1 and M 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 m e /1, 100 = 275.5 m e
It is a pion. Its accepted mass is 273 m e
7.40 Balancing the centripetal force with the magnetic force at the point of
extraction
mv
2
/r = zev B
(1)
which gives us
v = zeBr/m
The ratio of velocities for α-particle and deuteron is
v d /v α = (z d .m α )/(m d .z α ) = (1 × 2)/(2 × 1) = 1
since m α ≈ 2m d
As the initial velocities are identical the ratio
R d /R α = (m d z α
2 )/(m α z d
2 ) = 2
2
/2 = 2
Therefore, R d = 2R α
7.41 Use Geiger’s rule
R = K E
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
3/2
1
Or E 1 = 5.79 MeV
Differentiating (1)
dR/dE = (3/2)K E
1/2
dE 1 /dR = 2/3K E 1
1/2
= 2/(3 × 0.33 ×
√
5.79)
= 0.84 MeV/cm