The volume occupied by any lump of matter is due primarily to it’s atoms’ A ) electron clouds B ) protons C ) nuc

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

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

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

Earth Moon

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

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

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

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

82Pb207 w For a thin enough layer n(w)p(5  10-13cm)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!

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

We’ve named 3 forms of natural terrestrial radiation a b g How did these rank in ionizing power? in penetrability(range)? 1 2 3

We’ve named 3 forms of natural terrestrial radiation a b g How did these rank in ionizing power? in penetrability(range)? 1 2 3 3 2 1 Can you suggest WHY there is this inverse relationship between ionization and penetrability? “ionizing” radiation

mproton = 0.0000000000000000000000000016748 kg melectron=0.0000000000000000000000000000009 kg

Momentum depends on massandvelocity Easy to stop Hard to stop m v v m v Easy to stop Hard to stop v m m Momentumisinertia of motion While inertiadepends on mass Easy to start Hard to start “Quantity of motion” momentum = mass  velocity

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

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

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

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?

A100 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

v v mv mv (m+m)v mv mv

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 600 kg 25 m/sec 900 kg 10 m/sec B

A light particle of chargeq1encounters (passes by, not directly hitting) a heavy particle of chargeq2at rest, b q2 b “impact” parameter q1

A light particle of chargeq1encounters (passes by, not directly hitting) a heavy particle of chargeq2at rest, and follows a HYPERBOLIC TRAJECTORY b F' q2 F b “impact” parameter q1

A light particle of chargeq1encounters (passes by, not directly hitting) a heavy particle of chargeq2at 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 from and escaping back to distant space). q1

A light particle of chargeq1encounters (passes by, not directly hitting) a heavy particle of chargeq2at rest, and follows a HYPERBOLIC TRAJECTORY  b F´ q2 F  q1

  b q2 Largerdeflection m q v0 b smaller larger q1 much smaller smaller

Relaxing the “light”, “heavy” requirement simply means BOTH will move in response to the forces between them. 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

What about the ENERGY LOST in the collision? • 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)

A projectile with initial speed v0scatters 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

A projectile with initial speed v0scatters 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

mvf  mv0 mvf mv0

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

If scattering (  ) is small • large impact parameterb • and/or • large projectile speedv0 • vf  vo mvf  /2 p  /2 mv0 C Recall sin a = B/C B  A

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

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

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

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)

For nuclear collisions: mtarget 2Z2mproton For collisions with atomic electrons: mtarget melectronq2 = 1e for an encounter with 1 electron

For nuclear collisions: mtarget 2Z2mproton For collisions with atomic electrons: mtarget melectronq1 = 1e Z2 times as many of these occur!

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

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

energy loss speed

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) 101 102 104 105 106

The volume occupied by any lump of matter is due primarily to it’s atoms’ A ) electron clouds B ) protons C ) nuc