1. The volume occupied by
any lump of matter is due
primarily to it’s atoms’
A) electron clouds
B) protons
C) nuclei
D) other

2. The mass of matter is
due primarily to it’s
A) electron cloud
B) nuclei
C) other

3. If atoms are mostly empty space,
why don’t we just fall
through the floor?
A) electrical forces
B) magnetic forces
C) gravitational forces
D) nuclear forces
E) atoms are not
mostly empty space

4. Earth
Moon

5. Earth
Moon

6. for some sense of spacing consider the ratio orbital diameters
central body diameter
~ 10s for moons/planets
~100s for planets orbiting sun
In a solid
interatomic spacing: 1-5 Å (1-5  m)
nuclear radii: fm (1.5-5  m)
the ratio
orbital diameters
central body diameter
~ 66,666 for atomic electron
orbitals to their own nucleus
A basketball scale nucleus
would have its family of electrons stretching 10s of miles away

7. Carbon 6C
Oxygen 8O
Aluminum 13Al
Iron 26Fe
Copper 29Cu
Lead 82Pb
What about a single, high energy, charged particle?

8. n= rNA / A where NA = Avogadro’s Number
A solid sheet of lead offers how much of a (cross
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

9. n(Volume)  (atomic cross section) = n(surface area,A  w)(pr2)
82Pb207 w
For a thin enough layer
n(Volume)  (atomic cross section)
= n(surface area,A  w)(pr2)
as a fraction of the target’s area:
= n(w)p(5  10-13cm)2
5  10-15m

10. n(w)p(5  10-13cm)2 82Pb207 For 1 mm sheet of lead: 0.00257
For a thin enough layer
n(w)p(5  10-13cm)2
For 1 mm sheet of lead:
1 cm sheet of lead:

11. nw nuclei per unit area Actually a projectile “sees”
but Znw electrons per unit area!

12. a b g We’ve named 3 forms of natural terrestrial radiation
How did these rank
in ionizing power?

13. a b g We’ve named 3 forms of natural terrestrial radiation
How did these rank
in ionizing power?
in penetrability(range)?

14. a b g “ionizing” radiation
We’ve named 3 forms of natural terrestrial radiation
a b g
How did these rank
in ionizing power?
in penetrability(range)?
Can you suggest WHY there is this inverse relationship between ionization and penetrability?
“ionizing” radiation

15. mproton = kg
melectron= kg

16. Momentum is inertia of motion
While inertia depends on mass
Easy to start
Hard to start
Momentum
depends on mass and velocity
Easy to stop
Hard to stop v m
“Quantity of motion”
momentum = mass  velocity

17. Ft  Dp Impulse = force × time  Dp = F Dt To change velocity  Force
To change momentum  Impulse
Impulse = force × time  Dp = F Dt
examples
Ft  Dp
Ft = Dp
(doesn’t break)
(breaks)
NERF

18. A bowling ball and ping-pong
ball are rolling towards you
with the same momentum.
Which ball is moving toward
you with the greater speed?
A) the bowling ball
B) the ping pong ball
C) same speed for both

19. A) 0 (both stop). B) v/4 v/2 v v A fast moving car traveling with a
speed v rear-ends an identical model
(and total mass) car idling in neutral
at the intersection. They lock bumpers
on impact and move forward at
A) 0 (both stop).
B) v/4
v/2 v

20. A) right B) left A heavy truck and light car both
traveling at the speed limit v, have
a head-on collision. If they lock bumpers
on impact they skid together to the
A) right
B) left
Under what conditions would they stop dead?

21. A 100 kg astronaut at rest catches a
50 kg meteor moving toward him at
9 m/sec. If the astronaut manages
to hold onto the meteor after catching
it, what speed does he pick up?
A) 3 m/sec
B) 4.5 m/sec
C) 9 m/sec
D) 15 m/sec
E) 18 m/sec
F) some other speed

22. v v mv mv
(m+m)v mv mv

23. Car A has a mass of 900 kg and is travelling east at a speed of 10 m/sec. Car B has a mass of 600 kg and is travelling north at
a speed of 25 m/sec. The two cars collide, and lock bumpers. Neglecting friction which arrow best represents the direction the combined wreck travels? A B C A
900 kg
10 m/sec
600 kg
25 m/sec B

26. b q2 b “impact” parameter q1 A light particle of charge q1 encounters
(passes by, not directly hitting)
a heavy particle of charge q2 at rest, b q2
b “impact” parameter q1

27. b q2 b “impact” parameter q1 A light particle of charge q1 encounters
(passes by, not directly hitting)
a heavy particle of charge q2 at rest, b q2
b “impact” parameter q1

28. b q2 b “impact” parameter q1 A light particle of charge q1 encounters
(passes by, not directly hitting)
a heavy particle of charge q2 at rest, b q2
b “impact” parameter q1

29. b F' q2 F b “impact” parameter q1
A light particle of charge q1 encounters
(passes by, not directly hitting)
a heavy particle of charge q2 at rest,
and follows a HYPERBOLIC TRAJECTORY b F' q2 F
b “impact” parameter q1

30. F' b q2 F q1 A light particle of charge q1 encounters
(passes by, not directly hitting)
a heavy particle of charge q2 at rest,
and follows a HYPERBOLIC TRAJECTORY F' b q2 F
For an attractive “central” force the
heavy charge occupies the focus of
the trajectory like the sun does for a
comet sweeping past the sun (falling
fromand escaping back to distant space). q1

31.   b F´ q2 F q1 A light particle of charge q1 encounters
(passes by, not directly hitting)
a heavy particle of charge q2 at rest,
and follows a HYPERBOLIC TRAJECTORY  b F´ q2 F  q1

32.   Larger deflection m q v0 b b q2 q1 smaller larger much smaller

33. q2 Recoil of target q1 Relaxing the “light”, “heavy” requirement
simply means BOTH will move in
response to the forces between them. q2
Recoil of
target q1

34. q1 q2 Recoil of target q1 Relaxing the “light”, “heavy” requirement
simply means BOTH will move in
response to the forces between them. q1 q2
Recoil of
target q1

35. 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:
1/ (3672 Z)

36. A projectile with initial speed v0 scatters off
a target (as shown) with final speed vf.
mvf
mv0
The direction its target is sent recoiling
is best represented by
B C A T D E
G F

37. mvf mv0 F A projectile with initial speed v0 scatters off
a target (as shown) with final speed vf.
mvf
mv0
The sum of the final momentum (the
scattered projectile and the recoiling
target) must be the same as the initial
momentum of the projectile! F

38. mvf 
mv0
mvf
mv0

39. mvf 
mv0
( )
mvf
recoil
momentum
of target
(mv) = -
mv0

40. vf  vo If scattering (  ) is small large impact parameter b and/or
large projectile speed v0
vf  vo
mvf
 /2 p
 /2
mv0 C
Recall
sin a = B/C B  A

41. mvf
 /2 p
 /2
mv0 or
Together with:

42. q1 = Z1e q2 = Z2 e Recognizing that all charges are simple
multiples of the fundamental unit of the
electron charge e, we write
q1 = Z1e q2 = Z2 e

43. Z2≡Atomic Number, the number of protons (or electrons)
q2=Z2e
q1=Z1e

44. K = ½mv2 = (mv)2/(2m) K = (Dp)2/(2mtarget)
Recalling that kinetic energy
K = ½mv2 = (mv)2/(2m)
the transmitted kinetic energy
(the energy lost in collision to the target)
K = (Dp)2/(2mtarget)

45. mtarget  melectron q2 = 1e
For nuclear collisions: mtarget  2Z2mproton
For collisions with atomic electrons:
mtarget  melectron q2 = 1e
for an
encounter
with 1
electron

46. Z2 times as many of these occur! mtarget  melectron q1 = 1e
For nuclear collisions: mtarget  2Z2mproton
For collisions with atomic electrons:
mtarget  melectron q1 = 1e
Z2 times
as many
of these
occur!

47. The energy loss due to collisions with
electrons is GREATER by a factor of

48. Why are a-particles “more ionizing”
Notice this simple approximation
shows that
Why are a-particles “more ionizing”
than b-particles?

49. energy
loss
speed

50. r H2 gas target Pb target E (MeV) Felix Bloch Hans Bethe
-dE/dx = (4pNoz2e4/mev2)(Z/A)[ln{2mev2/I(1-b2)}-b2]
I = mean excitation (ionization) potential of atoms in target ~ Z10 eV
103
102
101
100
Range of dE/dx for proton through various materials
dE/dx ~ 1/b2
dE/d( x) r
H2 gas target
Pb target
Logarithmic rise
E (MeV)

51. g b Muon momentum [GeV/c] Particle Data Group, R.M. Barnett et al.,
Phys.Rev. D54 (1996) 1; Eur. Phys. J. C3 (1998)

52. m p a p d e D. R. Nygren, J. N. Marx, Physics Today 31 (1978) 46
dE/dx(keV/cm) e
Momentum [GeV/c]