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Calculation of classical formula of Stopping Power by Bohr
Hypothesis:
• The speed of the projectile >> speed of orbital electrons
– Atomic electron at rest
• The projectile is much heavier than the atomic electrons
• The projectile interacts through coulomb interactions only
– Nuclear reactions are not considered
• The interval of time where the interaction takes place is very small
– The atomic electron gets an impulse without changing position
• The projectile loses energy through interactions with orbital electrons
– Elastic and inelastic between the heavy charged particle and nuclei of the absorber are negligible
– The projectile trajectory is straight
Set-up
Fig1: Experimental set-up adopted in the calculation
Target material properties: Z, A, ρ
Projectile: Heavy charged particle with charge z and velocity v.
Calculation
∫∫ ⋅∆=⋅∆== bdbbEZA
NdbEN
dxdE
ddE Av
colle πσρξ
2)()(1, (1)
Where
,,2sec
,arg
,
parameterimpactbbdbtioncrossd
densityett
ZA
NN
Coll
Ave
===
=
=
πσρ
∆E(b) is the energy transferred to an orbital electron of the target, at a distance b from the
trajectory of the charged projectile (see Fig. 1):[ ]
embpbE
2)()(
2
=∆ (2).
vdxeEdteEFdtp ∫∫ ∫
+∞
∞−⊥
+∞
∞−⊥ ===∆ (3)
In (3) the parallel component of the electric field E gives null contribution for symmetry reasons.
Let’s apply the Gauss Theorem to the cylinder shown in Fig. 1.
∫∫∫+∞
∞−⊥
+∞
∞−⊥ =→=→=
bzedxEzebdxEzedsE
Cylinder 000 22
πεεπ
ε (4)
Let’s substitute (4) in (3) and we obtain:
bvze
bvze
vdxeEp
0
2
0
2
42
2 πεπε===∆ ∫
+∞
∞−⊥ (5)
Let’s substitute (5) in (2):
[ ]( ) ( ) 222
0
42
2220
422
42
424
2)()(
vbmez
vbmez
mbpbE
eee πεπε===∆ (6)
Let’s substitute (6) in (1), and we obtain (eq. 7):
( ) ( )
⋅=⋅=⋅∆= ∫∫
min
max2
2
20
4
220
42
ln4
414222)(
max
minbb
vz
meZ
AN
dbbvm
ezZA
NbdbbEZ
AN
ddE
e
Avb
be
AvAv
πεπ
πεππ
ξ
Let’s evaluate now bmin and bmax.
Let’s express b in function of ∆E(b) in equation (6).
,)(bE
Cb∆
=
Bmax corresponds to the minimum energy transfer possible, ∆Emin, equal to the mean excitation/ionization potential of the target material I.
Bmin corresponds to the maximum energy transfer, ∆Emax:
22
max 22
44vmMv
Mm
EMm
E ee
ke ===∆ , where Ek is the kinetic energy of the heavy charged
projectile.
We obtain that:
Ivm
EE
bb e
2
min
max
min
max 2== (8)
Finally, let’s substitute (8) in (7) and we obtain the classical formula of the Stopping Power by Bohr.
( ) ( )
⋅=
⋅=
Ivm
vz
meZ
AN
bb
vz
meZ
AN
ddE e
e
Av
e
Av2
2
2
20
4
min
max2
2
20
4 2ln
44ln
44
πεπ
πεπ
ξ