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Author Question: In the nuclear reaction of 12C( , n)15O, the probability of observing a neutron is much higher than ... (Read 163 times)

anjilletteb

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In the nuclear reaction of 12C( , n)15O, the probability of observing a neutron is much higher than observing the byproducts of a similar reaction such as 12C( , p)15N. Which of the following provides a possible explanation of this occurrence?
  a. There is always a higher probability for the compound nucleus to decay into products of lower mass.
  b. There is always a higher probability for the compound nucleus to decay into stable products.
  c. The compound nucleus 16O favors a decay to the n channel as opposed to the p channel for various reasons.
  d. The compound nucleus of 16O must make a forbidden transition to break into a proton and 15N.
  e. The entrance channel must enter into a compound nucleus to produce the 12C( , n)15O reaction as opposed to the direct reaction to produce the 12C( , p)15N reaction.

Question 2

Suppose in a dying star, a reaction takes place under extreme pressure and temperature. This reaction is hypothetically as follows:
 

 

a. Endoergic
  b. Exoergic


 

c. Elastic
  d. Break-even
  e. Fission


 



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Bison

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Answer to Question 1

c.
The existence of a resonance energy that the compound nucleus of 16O is excited to increases the probability of an exit channel that the quantum numbers of the nucleus select. In this case, the 12C( , n)15O reaction was selected and therefore it is more likely to observe the neutron.

Answer to Question 2

a.
The reaction is endoergic because of the mass energies associated with the products and reactants. The reaction takes in energy on the entrance channel of the equation to produce the exit channel. This reaction shows that after iron, the binding energy per nucleon decreases, meaning that any reaction fusing an element more massive than iron will perform an endoergic reaction.




anjilletteb

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Reply 2 on: Jul 28, 2018
Excellent


jamesnevil303

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Reply 3 on: Yesterday
Great answer, keep it coming :)

 

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