Wed. Jan. 14 – Physics Lecture #20
Relativity – Applications of the Relativistic Conservation Laws, Part 2
1) Applying Conservation Laws to “Relativistic Reactions”: decays/collisions
2) Fission and Fusion
3) A Star is Born, or,
Life, The Universe, and Everything (well, lots of stuff, anyway)
Warm-Up: Discuss with your neighbors
(Ch. R4, problem 4)
A particle of mass 3.0 MeV/ c 2 and momentum 1.0 MeV/ c hits and sticks to a particle of mass 2.0 MeV/ c 2 , initially at rest.
a) Find the mass of the composite resulting particle and its velocity.
b) How much kinetic energy is converted to mass?

A particle of mass 3.0 MeV/c 2 and momentum 1.0 MeV/c hits and sticks to a particle of mass 2.0 MeV/c 2 , initially at rest.
a) Find the mass of the resulting particle and its velocity.
b) How much kinetic energy is converted to mass?
before after u
2
= 0
3.0
p
1
= 1.0
2.0
m
3 u
3 m
1
 3 .
0 p
1
MeV / c 2  3 .
0
 1 .
0 MeV / c  1 .
0 m
2
 2 .
0 MeV / c 2  2 .
0 p
2
 0
E
2
 2 .
0 MeV  2 .
0 m
3 p
3
E
1
 10 MeV  10 E
3
E
1
 ( p
1 c ) 2  ( m
1 c 2 ) 2  ( p
Conserve momentum: p
1
 p
2
 p
3 p
3

1 .
0
1 .
0


0 
1 .
0 p
3
MeV / c
1
) 2  ( m
1
) 2  ( 1 .
0 ) 2
E
3

 ( 3 .
00 ) 2  10  E
1
 10 MeV
Conserve energy:
E
1
 E
2
 E
3
10
10  2  E
3
 2  5 .
162  5 .
162 MeV

p
3
 1 .
0 MeV / c
E
3
 5 .
162 MeV m
3

 m
3

( E
3
) 2
5 .
06
 ( p
3
) 2 
MeV / c 2
( 5 .
162 ) 2  ( 1 .
0 ) 2  5 .
06 u
3

 u
3 p
3
E
3


5 .
1 .
0
162
0 .
194 c
 0 .
194 m before
 m
 after
5 .
0 MeV, m after
 m before
 5 .
06
 5 .
06
MeV -
MeV
5 .
0 MeV
 kinetic energy converted to mass
 0.06
MeV
 0.06
MeV

2) Fission and Fusion
Chapter R4, problem 11)
In a nuclear reaction, two deuterium nuclei each of mass 1875.61 MeV/ c 2 combine to form a single helium nucleus of mass 3727.38 MeV/ c 2 .
Is rest energy converted to kinetic energy or vice-versa? Support your answer with a calculation.
Would you classify this as a fusion reaction or a fission reaction?

1.008
1.007
1.006
1.005
1.004
1.003
1.002
1.001
2
H
0.999
1
4
He
0.998
0
56 e
Fe
238
U
50 100
A
150 200 250
Figure 4.3.
Plot of average mass per nucleon (protons and neutrons) for the elements versus the number of nucleons A in the atom. The vertical axis has units amu which stand for atomic mass unit.

Concept Check: Based on the diagram, under which circumstances would it be possible to produce energy from matter? (Put up as many cards as are appropriate.)
1.008
2
H
1.007
1.006
1.005
1.004
1.003
1.002
1.001
0.999
1
4
He
0.998
0 50
56
Fe
1. Fuse two H atoms (A=2) together to get Helium (A=4)
100
A
150
2. Fuse two He atoms (A=4) together to get Berillium (A=8)
3. Fuse two Fe atoms (A=56) together to get Cadmium (A=112)
200
238
U
250
4. Fuse an atom of Uranium (A=238) with a Helium (A=4) and Hydrogen (A=2) to get Plutonium (A=244).

1937

1954

1976

Chapter R4, problem 15)
After a supermassive star has run out of Hydrogen to fuse, it starts fusing
Helium into heavier elements, then fusing those into heavier elements, etc., until it gets to iron (Fe). Up until this point, the fusion reactions produce kinetic energy in the form of rapidly moving atoms and molecules and photons. The star is effectively exploding, but its massive gravitational field keeps the explosion contained, and this balance results in a stable start star. But after the star has fused its materials into iron, it stops producing kinetic energy and the gravitational interaction causes it to contract very suddenly; the massive decrease in gravitational energy results in an uncontained explosion. This is one way in which a start star can go supernova.
Use Figure 4.3 to explain what is so special about when you try to fuse iron.
Why can’t the star produce additional kinetic energy via fusion of iron with itself or lighter elements?
1.008
2
H
1.007
1.006
1.005
1.004
1.003
1.002
1.001
0.999
1
4
He
0.998
0 50
56
Fe
100
A
150 200
238
U
250