8
1 Energy Release in Nuclear Reactions, Neutrons, Fission, and Characteristics …
must always be conserved. If the polonium nucleus is initially stationary, then the
lead and alpha nuclei must recoil in opposite directions. One can easily show from
classical momentum conservation that if the total kinetic energy shared by the two
product nuclei is Q, then the kinetic energy of the lighter product nucleus must be
K m =
Q
1 + m/M
,
(1.20)
where m and M are respectively the masses of the light and heavy product nuclei.
Here we have m/M ~ 4/206, so the alpha-particle carries off the lion’s share of the
liberated energy, about 5.3 MeV. The speed of such an alpha-particle is about 0.05c,
justifying the non-relativistic assumption.
We now set up some expressions that will be useful for dissecting Chadwick’s
analysis.
First, let us assume that Bothe & Becker and the Joliot-Curies were correct in
their interpretation that α-bombardment of Be creates gamma-rays. To conserve the
number of nucleons involved, they hypothesized that the reaction was
4
2 He +
9
4 Be →
13
6 C + γ.
(1.21)
(Strictly speaking, we are cheating here in writing the reaction in modern notation
that presumes knowledge of both neutrons and protons, but this has no effect on the
analysis.) From left to right, the -values for this reaction are 2.425, 11.348, and
3.125 MeV, so the Q-value is 10.65 MeV; this energy, when added to the ~ 5.3 MeV
kinetic energy of the incoming alpha, means that the γ -ray can have an energy of at
most about 16 MeV. However, the energy of the supposed gamma-ray is crucial here,
so we do a more careful analysis. In Appendix D, it is shown that if a collision like
this happens head-on and if the gamma-ray that is produced travels in the forward
direction after the reaction, the energy E γ of the emergent gamma-ray is given by
solving the quadratic equation
α E
2
γ + ε E γ + δ = 0,
(1.22)
where
α =
1
2E C
,
(1.23)
ε = 1 −
√
2E He K He
E C
,
(1.24)
and
δ =
E He
E C
K He − (E He + E Be + K He − E C ),
(1.25)
1 Energy Release in Nuclear Reactions, Neutrons, Fission, and Characteristics …
must always be conserved. If the polonium nucleus is initially stationary, then the
lead and alpha nuclei must recoil in opposite directions. One can easily show from
classical momentum conservation that if the total kinetic energy shared by the two
product nuclei is Q, then the kinetic energy of the lighter product nucleus must be
K m =
Q
1 + m/M
,
(1.20)
where m and M are respectively the masses of the light and heavy product nuclei.
Here we have m/M ~ 4/206, so the alpha-particle carries off the lion’s share of the
liberated energy, about 5.3 MeV. The speed of such an alpha-particle is about 0.05c,
justifying the non-relativistic assumption.
We now set up some expressions that will be useful for dissecting Chadwick’s
analysis.
First, let us assume that Bothe & Becker and the Joliot-Curies were correct in
their interpretation that α-bombardment of Be creates gamma-rays. To conserve the
number of nucleons involved, they hypothesized that the reaction was
4
2 He +
9
4 Be →
13
6 C + γ.
(1.21)
(Strictly speaking, we are cheating here in writing the reaction in modern notation
that presumes knowledge of both neutrons and protons, but this has no effect on the
analysis.) From left to right, the -values for this reaction are 2.425, 11.348, and
3.125 MeV, so the Q-value is 10.65 MeV; this energy, when added to the ~ 5.3 MeV
kinetic energy of the incoming alpha, means that the γ -ray can have an energy of at
most about 16 MeV. However, the energy of the supposed gamma-ray is crucial here,
so we do a more careful analysis. In Appendix D, it is shown that if a collision like
this happens head-on and if the gamma-ray that is produced travels in the forward
direction after the reaction, the energy E γ of the emergent gamma-ray is given by
solving the quadratic equation
α E
2
γ + ε E γ + δ = 0,
(1.22)
where
α =
1
2E C
,
(1.23)
ε = 1 −
√
2E He K He
E C
,
(1.24)
and
δ =
E He
E C
K He − (E He + E Be + K He − E C ),
(1.25)