1 / 20

# v D t - PowerPoint PPT Presentation

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' v D t' - benjamin-stewart

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Incident mono-energetic beam

v Dt

A

dW

N = number density in beam

(particles per unit volume)

Solid angle dWrepresents

detector counting the dN

particles per unit time that

scatter through qinto dW

Nnumber of scattering

centers in target

intercepted by beamspot

FLUX = # of particles crossing through unit cross section per sec

= NvDt A / Dt A = Nv

Notice: qNv we call current, I, measured in Coulombs.

dN NF dW dN = s(q)NF dW dN =NFds

-

dN = FNs(q)dWNFds(q)

the “differential” cross section

R

R

R

R

R

R

the differential solid angle d for integration is sin d d

Rsind

Rd

Rsind

Rd

Rsin

and consider

rings defined

by  alone

R

Rsind

R

R

R

Nscattered= NFsTOTAL

Integrated over all solid angles

Nscattered= NFsTOTAL

The scattering rate

per unit time

Particles IN (per unit time) = FArea(ofbeamspot)

Particles scattered OUT (per unit time) = F NsTOTAL

Moon

Moon

orbital diameters

central body diameter

~ 10s for moons/planets

~100s for planets orbiting sun

• In a solid

• interatomic spacing:1-5 Å (1-5  10-10 m)

• nuclear radii: 1.5 -5 f(1.5-5  10-15 m)

the ratio

orbital diameters

central body diameter

~ 66,666 for atomic electron

orbitals to their own nucleus

Carbon 6C

Oxygen 8O

Aluminum 13Al

Iron 26Fe

Copper 29Cu

sectional) physical target (and how much empty

space) to a subatomic projectile?

82Pb207

w

Number density,n: number of individual

atoms (or scattering centers!) per unit volume

n= rNA / A where NA = Avogadro’s Number

A = atomic weight (g)

r = density (g/cc)

n= (11.3 g/cc)(6.021023/mole)/(207.2 g/mole)

= 3.28  1022/cm3

82Pb207

w

For a thin enough layer

n(Volume)  (atomic cross section)

= n(surface areaw)(pr2)

as a fraction of the target’s area:

= n(w)p(5  10-13cm)2

For 1 mm sheet of lead: 0.00257

1 cm sheet of lead: 0.0257

nw nuclei per unit area

but Znw electrons per unit area!

let’s augmented with the specific example of

Coulomb scattering

will move in response to

the forces between them.

q1

q2

Recoil of

target

But here we are

interested only

in the scattered

projectile

q1

b

q2

d

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 d

d = 2 b db

using

becomes:

b

q2

d

and

so

b

q2

d

• the recoiling target carries energy

• some of the projectile’s energy was surrendered

• if the target isheavy

• the recoil is small

• the energy loss is insignificant

Reminder:

1/ (3672 Z)