lecture 7: matter and the four forces
TRANSCRIPT
Lecture 7: Matter and the Four Forces
atomic nuclei1
1p =1
1 H
2
1H = deuterium
3
1H = tritium
stable, mass less than p + n (more binding E)
unstable
atomic nucleiwhy doesn’t deuterium decay like free neutron?
Pauli’s exclusion principle (neutron can’t decay, proton already in that state)!
3
1H →3
2 He +0
−1 e +0
0 ν
atoms are most stable near A-Z ~ Z, where number of neutrons, protons ~same
atomic weight A
atomic number Z
tritium unstable for same reason!
Thermonuclear Reactions1
1H +1
1 H →2
1 H +0
1 e +0
0 ν
indicates weak interaction
formation of deuterium
need to overcome Coulomb barrier (strong nuclear force short-range, EM long-range): high T + quantum tunneling
fewer higher energy particles, but better at tunneling
high core T ---> more high energy particles
net probability of coming within nuclear distance proportional to product of two competing exponentials
probability of relative speed v ∝ exp(−mv2/2kT )
penetration probability
probability of relative speed
∝ exp(−4π2q1q2/hυ)
Thermonuclear Reactions
1
1H +1
1 H →2
1 H +0
1 e +0
0 ν
indicates weak interaction
liberated energy carried off as increased kinetic energy of reaction products, more heating, but neutrinos escape quickly
positron eventually annihilates, contributes more energy
reaction slow, low probability due to weak interaction
formation of deuterium
Proton-Proton Chain1
1H +1
1 H →2
1 H +0
1 e +0
0 ν
2
1H +1
1 H →3
2 He +0
0 γ
32He +
32 He →
42 He +
11 H +
11 H
first step gives chain its name
= gamma ray released0
0γ
gamma ray needed to conserve energy & linear momentum
why isn’t it easier to use 3
2He +1
1 H ?
steps 1 & 2 happen twice
neutrino correction to total energy produced
what if first step wasn’t mediated by weak interaction?
central T (15 million K!) higher than P-P chain requires: are there other possible reaction chains for making He?
C-N-O cycle126 C +
11 H →
137 N +
00 γ
137 N →
136 C +
01 e +
00 ν
136 C +
11 H →
147 N +
00 γ
147 N +
11 H →
158 O +
00 ν
158 O →
157 N +
01 e +
00 ν
157 N +
11 H →
126 C +
42 He
C, N, O are catalysts, net abundances unchanged
why don’t two beta decays make this improbable like in p-p chain?
sequence is cyclic, we could start anywhere with proton + C, N, or O870 s
178 s
advantageous for H ---> He, what about He ---> heavier nuclei?
key is binding energy per nucleon (or nuclear baryon)
extract energy by fusion process indefinitely?
increasing binding E up to He-4 shows why p-p chain has direction,exothermic reactions
binding energy per baryon ~increases until Fe
conflict between short-range strong force and EM
if nucleus small, strong force wins over EM, still advantageous to add another baryon
heavier nuclei physically bigger, at some point, ie, Fe, EM~strong force, adding more baryons causes loss of binding energy
2) adding neutron destabilizes nucleus, breaks into ~2 equal pieces (fission), releases more free neutrons in chain reaction
3) producing Fe is end of line; once (high-mass) star makes Fe, no further reactions can release energy ---> BIG trouble
4) trend is general, not monotonic: next stable nucleus after He-4 is Li-6, less binding energy/nucleon (even # of protons, neutrons more stable, can stack 2 at time, with opposite spins)
How does nature surpass He-4 barrier? Or make elements beyond Fe?
1) very heavy nuclei tend to split, alpha decay = He-4 nucleus
92U most massive, naturally-occurring element on Earth,alpha, beta decays lead to lead
consequences...
going beyond He-4...
Triple Alpha Reaction42He +
42 He →
84 Be +
00 γ
Be decays (back to 2 He) in 2.6 x 10^-16 sec!!!!
if T = 10^8 K and density = 10^5 gm/cm^3, one Be nuclei for every 10^9 He in equilibrium
84Be +
42 He →
126 C∗
+00 γ
in other words, three He-4 make C and two gamma rays
indicates excited nuclear state, predicted by Hoyle before excited nucleus discovered!
other leapfrogging then happens ...
126 C +
42 He →
168 O +
00 γ
126 C +
126 C →
2412 Mg +
00 γ
heavier elements generally have larger charges
fusion then requires overcoming ever larger Coulomb barriers
burning heavier elements generally requires larger temperatures,star’s evolution brings these higher T’s