10
1 Energy Release in Nuclear Reactions, Neutrons, Fission, and Characteristics …
E γ again designates the energy of the gamma-ray before the collision. Maximum
possible forward momentum will be imparted to the struck particle if the gammaray recoils backwards after the collision, so we assume that this is the case. If the
energy of the gamma-ray after the collision is E γ *, then conservation of mass-energy
demands
E γ + E m = E
∗
γ + E m + K m .
(1.26)
Since we are assuming that the struck particle does not change its identity, the
factors of E m in (1.26) cancel each other. Since the momentum of a photon of energy
E is given by E/c, conservation of momentum for this collision can be written as
E γ /c = −E
∗
γ /c + mv,
(1.27)
where v is the post-collision speed of the struck particle. The negative sign in the
right side of (1.27) means that the γ -ray recoils backwards.
It will be useful to also have on hand expressions for the classical momentum and
kinetic energy of the struck particle in terms of its rest energy:
mv =
2m K m =
2mc 2 K m
c
=
√
2E m K m
c
(1.28)
and
K m =
1
2
mv
2
=
mc
2
v
2
2c 2
=
1
2
E m
v
c
2 .
(1.29)
With (1.28), a factor of c can be cancelled in (1.27); then, on eliminating E γ *
between (1.26) and (1.27), we can solve for E γ :
E γ =
1
2
K m +
2E m K m
.
(1.30)
If the kinetic energy of the struck particle (proton) can be measured, we can use
(1.30) to determine what energy the gamma-ray must have had to set it into such
motion. On the other hand, if we desire to solve for K m presuming that E γ is known,
the situation is messier as (1.30) is a quadratic in
√
K m that has no neat solution:
K m +
2E m
K m − 2E γ = 0.
(1.31)
We will need an expression for K m , however, so we must solve the quadratic.
Fortunately, we can invoke a simplifying approximation.
The gamma-rays involved here have E γ ~ 14.6 MeV, but a proton has a rest energy
of about 938 MeV. It is consequently quite reasonable to set E m >> E γ , in which case
this quadratic can be solved approximately, as shown in what follows.
1 Energy Release in Nuclear Reactions, Neutrons, Fission, and Characteristics …
E γ again designates the energy of the gamma-ray before the collision. Maximum
possible forward momentum will be imparted to the struck particle if the gammaray recoils backwards after the collision, so we assume that this is the case. If the
energy of the gamma-ray after the collision is E γ *, then conservation of mass-energy
demands
E γ + E m = E
∗
γ + E m + K m .
(1.26)
Since we are assuming that the struck particle does not change its identity, the
factors of E m in (1.26) cancel each other. Since the momentum of a photon of energy
E is given by E/c, conservation of momentum for this collision can be written as
E γ /c = −E
∗
γ /c + mv,
(1.27)
where v is the post-collision speed of the struck particle. The negative sign in the
right side of (1.27) means that the γ -ray recoils backwards.
It will be useful to also have on hand expressions for the classical momentum and
kinetic energy of the struck particle in terms of its rest energy:
mv =
2m K m =
2mc 2 K m
c
=
√
2E m K m
c
(1.28)
and
K m =
1
2
mv
2
=
mc
2
v
2
2c 2
=
1
2
E m
v
c
2 .
(1.29)
With (1.28), a factor of c can be cancelled in (1.27); then, on eliminating E γ *
between (1.26) and (1.27), we can solve for E γ :
E γ =
1
2
K m +
2E m K m
.
(1.30)
If the kinetic energy of the struck particle (proton) can be measured, we can use
(1.30) to determine what energy the gamma-ray must have had to set it into such
motion. On the other hand, if we desire to solve for K m presuming that E γ is known,
the situation is messier as (1.30) is a quadratic in
√
K m that has no neat solution:
K m +
2E m
K m − 2E γ = 0.
(1.31)
We will need an expression for K m , however, so we must solve the quadratic.
Fortunately, we can invoke a simplifying approximation.
The gamma-rays involved here have E γ ~ 14.6 MeV, but a proton has a rest energy
of about 938 MeV. It is consequently quite reasonable to set E m >> E γ , in which case
this quadratic can be solved approximately, as shown in what follows.