Coulomb potential in nuclear fission? In the case of symmetrical fission we must consider the Coulo

britesoulusjhq 2022-05-10 Answered

Coulomb potential in nuclear fission?
In the case of symmetrical fission we must consider the Coulomb potential between the two nuclei :

E = k \frac{q^{2}}{r} = k \frac{(\frac{Z e}{2})^{2}}{2 r_{0} (\frac{a}{2})^{1 / 3}} = k \frac{2^{1 / 3}}{8} \frac{Z^{2} e^{2}}{r_{0} A^{1 / 3}} \frac{1}{1.6 \times 10^{- 13}} = 0.157 \frac{Z^{2}}{A}

The process is spontaneous only if

Q > E

(where

Q

is the difference between final and initial binding energy).
Why is the process allowed only if

Q > E

? Don't the nuclei repel each other rendering the process "easier"?

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Maeve Holloway

Answered 2022-05-11 Author has 25 answers

You might want to cite the source of the quote.
"Don't the nuclei repel each other rendering the process "easier"?"
The actual potential is a sum of nuclear and electrical potentials. The sum has a barrier with the height they approximated in this quote. (Their calculation is only an approximation, because they didn't actually include the nuclear potential.) Decay requires quantum-mechanical tunneling through this barrier. The tunneling probability doesn't have anything to do with whether the force is attractive or repulsive, it depends on the height and width of the barrier. By definition, any barrier is repulsive on one side and attractive on the other, because the derivative

V^{'}

flips signs. If you want to see more details, and sketches of graphs, see the example titled "Tunneling in alpha decay" in my book Simple Nature, ca. p. 889. Alpha decay can be considered as a type of very asymmetric fission.
Why is the process allowed only if

Q > E

?
This is just conservation of energy. The

Q

value is the energy that is released by the process. The energy

E

is the electrical potential energy at the point of scission (or, more accurately, at the point where the nuclear force becomes negligible). Almost all of this energy

E

is then converted into the kinetic energy of the fragments. (Some also comes out in other forms, such as neutrons.)
Jon Custer said in a comment:
"If there is not enough energy to be able to separate the nucleus in the first place, to make two nuclei repelling each other, then you can never get that separation - the nucleus is energetically more stable as a bound state."
Not true. The process is classically forbidden, so we are guaranteed that there is not enough energy to separate the nuclei.

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