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444
Chapter 44
Each incident proton has half of the rest mass energy, or 1210 MeV = 1.21 GeV.
EVALUATE:
As we saw in problem 44.10, we would need much more initial energy if one of the initial protons
were stationary. The result here (1.21 GeV) is the
minimum
amount of energy needed; the original protons could
have more energy and still trigger this reaction.
44.17.
Section 44.3 says
02
(Z )
91.2 GeV
.
mc
=
98
2
2
5
0
91.2 10 eV
1.461 10
J;
1.63 10
kg; (Z )
(p)
97.2
Em
E
c
m
m
−−
=×
=
×
=
=
44.18.
(a)
We shall assume that the kinetic energy of the
0
Λ
is negligible. In that case we
can set the value of the photon’s energy equal to
Q
:
photon
(1193 1116) MeV
77 MeV
.
QE
=−
=
=
(b)
The momentum of this photon is
61
8
photon
20
8
(77 10 eV)(1.60 10
J eV)
4.1 10
kg m s
(3.00 10 m s)
E
p
c
−
−
××
==
=
×
⋅
×
To justify our original assumption, we can calculate the kinetic energy of a
0
Λ
that has this value of momentum
0
22
2
2
(77 MeV)
2.7 MeV
77 MeV.
2
2
2(1116 MeV)
pE
KQ
mm
c
Λ
=
=
<
<
=
Thus, we can ignore the momentum of the
0
Λ
without introducing a large error.
44.19.
IDENTIFY
and
SET UP:
Find the energy equivalent of the mass decrease.
EXECUTE:
The mass decrease is
0
() (
p
) ()
m
π
+
∑
and the energy released is
2
0
(
)
(p)
(
) 1189 MeV
938.3 MeV 135.0 MeV
116 MeV.
mc
mc
mc
+
∑
=
−
−
=
(The
2
mc
values for each
particle were taken from Table 44.3.)
EVALUATE:
The mass of the decay products is less than the mass of the original particle, so the decay is
energetically allowed and energy is released.
44.20.
IDENTIFY:
If the initial and final rest mass energies were equal, there would be no left over energy for kinetic
energy. Therefore the kinetic energy of the products is the difference between the mass energy of the initial
particles and the final particles.
SET UP:
The difference in mass is
0
K
mM m m
ΩΛ
Δ=
−
−
.
EXECUTE:
Using Table 44.3, the energy difference is
2
(
)
1672 MeV 1116 MeV
494 MeV
62 MeV
c
=Δ
=
−
−
=
EVALUATE:
There is less rest mass energy after the reaction than before because 62 MeV of the initial energy
was converted to kinetic energy of the products.
44.21.
Conservation of lepton number.
(a)
ee
e:
1
1
,
:
0
1
1
vv L
L
μμ
μ
→++
⇒
+≠−
≠++
, so lepton numbers are not conserved.
(b)
0
1
1
τ
τ
⇒
=+ −
;
: 1
1
τ
L
+=+
, so lepton numbers are conserved.
(c)
e.
γ
++
→+
Lepton numbers are not conserved since just one lepton is produced from zero original leptons.
(d)
np
e
γ
:0
1 1,
L
−
→+ +
⇒
so the lepton numbers are conserved.
44.22.
IDENTIFY
and
SET UP:
p and n have baryon number
+1 and
p has baryon number
1
−
.
e
+
, e
−
,
e
υ
and
all
have baryon number zero.
Baryon number is conserved if the total baryon number of the products equals the total
baryon number of the reactants.
EXECUTE: (a)
reactants:
1 1
2
B
=+=
.
Products:
101
B
=+ =
.
Not conserved.
(b)
reactants:
1 1
2
B
.
Products:
0
0
0
B
.
Not conserved.
(c)
reactants:
1
B
=+
.
Products:
1 0 0
1
B
=+ + =+
.
Conserved.
(d)
reactants:
1 1
0
B
=−=
.
Products:
0
B
=
.
Conserved.
44.23.
IDENTIFY
and
SET UP:
Compare the sum of the strangeness quantum numbers for the particles on each side of
the decay equation. The strangeness quantum numbers for each particle are given Table 44.3.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Energy, Mass

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