How many neutrons are produced in nuclear fission?

[songoku]

Homework Statement

When $U_{92}^{235}$ is bombarded by neutrons, nuclear fission can occur.
In one possible fission reaction, the two nuclides $Ba_{56}^{141}$ and $Kr_{36}^{92}$
are produced. How many neutrons are produced?
a. 1
b. 2
c. 3
d. 4
e. 5

Relevant Equations

Not sure

The nuclear fission reaction is:

$U_{92}^{235} + n^{1}_{0} \to Ba_{56}^{141} +Kr_{36}^{92} + 3 ~n^{1}_{0}$

I am not sure about the number of neutrons produced. Is it 2 or 3?

2 is from the number of neutrons on the RHS - number of neutron on LHS

3 is from the number of neutrons on RHS only

My guess would be 3. Is this correct? Thanks

 

[Delta2]

I am not sure either but I think its 2, because the question asks for the number of neutrons produced by the reaction. 1 Neutron is not produced, it is rather there from the start of the reaction.

 

[songoku]

Thank you very much Delta2

 

[Steve4Physics]

Homework Statement:: When $U_{92}^{235}$ is bombarded by neutrons, nuclear fission can occur.
In one possible fission reaction, the two nuclides $Ba_{56}^{141}$ and $Kr_{36}^{92}$
are produced. How many neutrons are produced?
a. 1
b. 2
c. 3
d. 4
e. 5
Relevant Equations:: Not sure

The nuclear fission reaction is:

$U_{92}^{235} + n^{1}_{0} \to Ba_{56}^{141} +Kr_{36}^{92} + 3 ~n^{1}_{0}$

I am not sure about the number of neutrons produced. Is it 2 or 3?

2 is from the number of neutrons on the RHS - number of neutron on LHS

3 is from the number of neutrons on RHS only

My guess would be 3. Is this correct? Thanks

The incident neutron is absorbed by the U-235 nucleus making (very unstable) U-236. This very quickly splits into fragments and, in this case, 3 neutrons.

Note - the 3 neutrons do not necessarily include the original incident neutron.

So I would go for answer c), 3 neutrons.

 

[Delta2]

Not good in nuclear physics here, but why the U-236 doesn't give back a U-235 and the 1 neutron?(some sort of reversible process i could say)... I guess the answer to this might be very complex...

 

[jbriggs444]

Not good in nuclear physics here, but why the U-236 doesn't give back a U-235 and the 1 neutron?(some sort of reversible process i could say)... I guess the answer to this might be very complex...

It doesn't happen because the problem stipulates that a different process happens instead.

But could it happen... Let us look up the atomic mass of U-235, U-236 and the atomic mass of a free neutron.

U-235: 235.043929
Neutron: 1.00866491588
Total: 236.0526
U-236: 236.04557

That means that neutron capture by U-235 yielding U-236 results in a decrease in total mass and, consequently a release of energy. You are going to end with the U-236 in a high energy state.

I am not sure whether, from this activated state, the U-236 nucleus can re-emit a neutron and whether this result is distinguishable from a more ordinary deflection of the neutron by the original U-235 nucleus.

If we look at the Wiki article on U-236 we find that...

When 235U absorbs a thermal neutron, one of two processes can occur. About 82% of the time, it will fission; about 18% of the time, it will not fission, instead emitting gamma radiation and yielding 236U.

"emitting gamma radiation" is a strong hint that the neutron capture is exothermic.

If you want U-236 to decay into U-235 plus a neutron, you will have to provide energy somehow.

 

[Steve4Physics]

Not good in nuclear physics here, but why the U-236 doesn't give back a U-235 and the 1 neutron?(some sort of reversible process i could say)... I guess the answer to this might be very complex...

I’m no expert either. But my understanding is this...

If the neutron is fast, then it will effectively pass through the U-235 without absorption. To get fission you need slow neutrons to increase the probability of absorption. Hence the use of a moderator.

When a slow neutron is absorbed, some energy (equal to the change in binding energy associated with strong forces between nucleons) is released. This leaves the (new) U-236 nucleus in an excited state.

There are various decay-modes for the excited U-236 nucleus. Releasing a neutron and leaving a U-235 nucleus is only one possibility.

But other decay-modes can occur. The excited U-236 nucleus can split into fragments + free neutrons in various possible ways. The reaction in the original question is one of many possibilities and is the one we are asked about.

Which decay-mode occur is governed by quantum mechanics so is effectively random. I’d guess that in general, the higher probability decay modes correspond to the larger potential energy losses (largest increases in binding energy per nucleon).

The decay-modes corresponding to the biggest energy losses produce the most stable products. That’s why you tend to end up with products (isotopes) having large binding energies per nucleon (high stability).

Releasing a neutron and leaving a U-235 nucleus may occur (not sure of the probability). But the other decay-modes (such as the one in the original question) also occur.

If that helps!

 

[jbriggs444]

I’m no expert either. But my understanding is this...

If the neutron is fast, then it will effectively pass through the U-235 without absorption. To get fission you need slow neutrons to increase the probability of absorption. Hence the use of a moderator.

Same "not an expert" caveat.

This would fit with an idea that absorbing a neutron puts the resulting nucleus in a high energy state. If no state of sufficient energy exists, the capture cannot happen.

The states of nucleons in the nucleus are apparently analogous to the states of electrons in an atom -- there are discrete energy levels.

 

[kuruman]

$U_{92}^{235} + n^{1}_{0} \to Ba_{56}^{141} +Kr_{36}^{92} + 3 ~n^{1}_{0}$

I think it is a matter of semantics about the meaning of "produced".

We are given a specific reaction and the question is "How many neutrons are produced?" Could this mean "What is the difference in number of neutrons, $\Delta N = N_{\text{after}}-N_{\text{before}}$? Then
$N_{\text{before}}=(235-92)+1=144$
$N_{\text{after}}=(141-56)+(92-36)+3=144$
Therefore $\Delta N=0$ but that's none of the choices.

In fact a neutron is meaningfully "produced" in electron capture whereby a nuclear proton is converted into a neutron via the weak interaction, but that's not what we have here. So bound neutrons don't count and we should consider only the unbound neutrons. Well, we could argue that if one of the three "after" neutrons is the same particle as the one before, then two neutrons are produced. However, neutrons are indistinguishable particles and the argument works only with distinguishable particles. I guess this leaves 3 as the answer. BTW, I am not an expert either.

 

[songoku]

Since everyone is not an expert, so it leaves only me as the expert...

Thank you very much for the help and explanation Delta2, Steve4Physics, jbriggs444, kuruman