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