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scattered particles. Incident mono-energetic beam. v D t. A. d W. N = number density in beam (particles per unit volume). Solid angle d W represents detector counting the dN particles per unit time that scatter through q into d W. N number of scattering. centers in target - PowerPoint PPT Presentation
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v t
d
Incident mono-energetic beamscattered particles
A
N = number density in beam (particles per unit volume)
N number of scattering centers in targetintercepted by beamspot
Solid angle d representsdetector counting the dN particles per unit time that
scatter through into d
FLUX = # of particles crossing through unit cross section per sec = Nv t A / t A = Nv
Notice: qNv we call current, I, measured in Coulombs.
dN N F d dN = N F d dN = N F d
dN = FN d N F d
the “differential” cross section
R
R
R
R
R
the differential solid angle d for integration is sin dd
R
R
Rsin
RsindRd
RsindRd
ddR
ddRd sin
sin2
2
Symmetry arguments allow us to immediately integrate out
Rsind
R
RR
R
and consider rings definedby alone
Integrated over all solid angles Nscattered = N F TOTAL
Nscattered = N F TOTALThe scattering rate
per unit time
Particles IN (per unit time) = FArea(of beam spot)
Particles scattered OUT (per unit time) = F N TOTAL
AvogadroN
A
N
Earth Moon
Earth Moon
In a solid•interatomic spacing: 15 Å (15 10-10 m)•nuclear radii: 1.5 5 f (1.55 10-15 m)
for some sense of spacing consider the ratio
orbital diameterscentral body diameter
~ 10s for moons/planets
~100s for planets orbiting sun
the ratio orbital diameterscentral body diameter
~ 66,666 for atomic electronorbitals to their own nucleus
Carbon 6COxygen 8OAluminum 13AlIron 26FeCopper 29CuLead 82Pb
A solid sheet of lead offers how much of a (cross sectional) physical target (and how much empty space) to a subatomic projectile?
82Pb207
Number density, n: number of individual atoms (or scattering centers!) per unit volume
n= NA / A where NA = Avogadro’s Number
A = atomic weight (g) = density (g/cc)
w
n= (11.3 g/cc)(6.021023/mole)/(207.2 g/mole)
= 3.28 1022/cm3
82Pb207w
For a thin enough layer
n(Volume) (atomic cross section)= n(surface areaw)(r2)
as a fraction of the target’s area: = n(w)13cm)2
For 1 mm sheet of lead: 0.00257 1 cm sheet of lead: 0.0257
Actually a projectile “sees”
nw nuclei per unit area
but Znw electrons per unit area!
that general description of cross section
let’s augmented with the specific example of
Coulomb scattering
q2
Recoil oftarget
BOTH target and projectile will move in response to
the forces between them. q1
q1
20
21202
tanbmv
bmv
K
But here we areinterested onlyin the scattered
projectile
impact parameter, b
d
q2
b
A beam of N incident particles strike a (thin foil) target.The beam spot (cross section of the beam) illuminates n scattering centers.
If dN counts the average number of particles scattered between and d
dN/N = n dusing
dx
du
uu
dx
d2cos
1tan
20
21
2tan
mbv
becomes:
dbvmb
qqd 2
02
21
2
2cos2
1
d = 2 b db
d
q2
b
20
21
2tan
mbv
2tan2
0
21
mv
qqb
dbvmb
qqd 2
02
212 2cos2
1
and
dqq
vmbdb
2cos2 221
20
2
so
d
mv
mvd
3
20
212
21
20
2tan2cos
d
mv
mvd
3
20
212
21
20
2tan2cos
sin
2sin2
2cos2sin
2sin
2cos
2sin
2cos
44
3
32
d
mv
qqd
2sin2
sin4
2
20
21
d
q2
b
What about the ENERGY LOST in the collision?
•the recoiling target carries energy•some of the projectile’s energy was surrendered•if the target is heavy
•the recoil is small•the energy loss is insignificant
Reminder:
proton
e
proton
e
nucleus
e
Zm
m
Am
m
m
m
2
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