What Is the Q-Value of Pu-239 Fission into Sr-92, He-4, and Neutrons?

[says]

Homework Statement

Fission Reaction: Pu-239 + gamma --> Sr-92 + ? + 3n

Q-Value = ? (Answer in MeV, correct to 5 significant figures)

Atomic Masses (u)
Alpha Particle = 4.00150618
He-4 = 4.0026032497
Pu-239 = 239.0521565
Sr-92 = 91.911030
n (neutron) = 1.00866501

Homework Equations

Q-Value = Q = E = Δmc² = (m_i - m_f)c²

The Attempt at a Solution

Q = (239.0521565 + gamma) - (91.911030 + 4.0026032497 + (3*1.00866501)) * c²
= (239.0521565 + gamma) - (98.939628)
Q = gamma = -140.1125285 * 931.5020
= -130515.1
= - 130520 MeV (correct to 5 significant figures)

Two Questions
1) I've seen examples where He-4 = 4.0026032497 is used in this calculation instead of the Alpha Particle = 4.00150618. Why is this? I know He-4 is missing it's two electrons, therefore it's mass is the equivalent of the alpha particle, but how come He-4 atomic mass ≠ alpha particle. And why should I use the atomic mass of He-4 in this calculation instead of the atomic mass of the alpha particle?

2) Is Q negative because it represents the amount of energy which must be provided for the reaction to occur? I.e. A gamma ray with 130520 MeV must be provided to split a Pu-239 atom into Sr-92 + He-4 + 3n

 

[atomicpedals]

I'm a bit confused as to why you've introduced an alpha into your solution? Looking at the problem statement there are only the Pu, gamma, Sr, and 3n.

 

[says]

Pu-239 decays via alpha decay.

so Pu-239 + gamma --> Sr-92 + ? + 3n

If Pu-239 decays via alpha decay then ' ? ' = He-4

Pu-239 + gamma --> Sr-92 + He-4 + 3n

Sorry, that was the first part of the question but I figured that out from a quick google search.

 

[atomicpedals]

To answer your question on the mass difference between He-4 and an alpha, He-4 has two electrons (which are on the order of 0.0005amu each) and an alpha has none - an alpha is just a He-4 nucleus.

 

[atomicpedals]

Your Q value should be positive, it is the amount of energy released from the reaction. When I read the question as set up what I see is gamma induced fission of Pu resulting in Sr, 3n, and a lot of energy. The fact that Pu-239 decays via alpha emission doesn't factor in. So what I would set up to solve is ((mass Pu)-(mass Sr + mass 3n))c^2

 

[says]

Sorry, the full question states:

Complete the following fission reaction, and then calculate the Q-value for the reaction: Pu-239 + gamma --> Sr-92 + ? + 3n

I'm pretty sure that means Pu-239 + gamma --> Sr-92 + He-4 + 3n

I thought because the energy of the gamma ray is the only unknown then Q = energy of the gamma ray = 130520 MeV

Because the energy of the gamma ray is the amount of energy that has to be provided for the reaction of Pu-239 + gamma --> Sr-92 + He-4 + 3n to occur, this means gamma = -130520 MeV

 

[Astronuc]

Pu-239 decays via alpha decay.

so Pu-239 + gamma --> Sr-92 + ? + 3n

If Pu-239 decays via alpha decay then ' ? ' = He-4

Pu-239 + gamma --> Sr-92 + He-4 + 3n

Sorry, that was the first part of the question but I figured that out from a quick google search.

The OP indicates gamma-induced fission (as opposed to neutron-induced fission). The ? should be a fission product, not an alpha particle. The sums of Zs and As on the leftside of the equation should equal the sum of the Zs and As. The gamma ray has no charge or mass. Z(Pu) = 94, Z(Sr) = 38. Neutrons have no charge, but atomic mass, A = 1.

 

[atomicpedals]

The Q value is a measure of the energy released in a reaction. So in this case it is the energy released in a Pu+gamma reaction (gamma induced fission of Pu). A positive Q value means the reaction is exothermic (which fission very much is), a negative would indicate the reaction was endothermic.

 

[SteamKing]

Sorry, the full question states:

Complete the following fission reaction, and then calculate the Q-value for the reaction: Pu-239 + gamma --> Sr-92 + ? + 3n

I'm pretty sure that means Pu-239 + gamma --> Sr-92 + He-4 + 3n

I thought because the energy of the gamma ray is the only unknown then Q = energy of the gamma ray = 130520 MeV

Because the energy of the gamma ray is the amount of energy that has to be provided for the reaction of Pu-239 + gamma --> Sr-92 + He-4 + 3n to occur, this means gamma = -130520 MeV

This is why the PF HW forum rules ask posters to include the full question when filling out the HW template. Otherwise, a lot of time is wasted in going back and forth trying to make sense of what is being asked.

Also, the fission reaction:

Pu-239 + gamma --> Sr-92 + He-4 + 3n

needs examination. Has anyone tried to balance the left hand side with the right hand side? What's the atomic number of plutonium? Of strontium?

A gamma ray with an energy of 130,000 MeV is an extraordinary event. You only get about 200 MeV from the fission of a single atom of U-235.

 

[says]

Pu-239 ---> Sr-92 + Ba-147 + 3n

Z
Pu = 94
Sr = 38
Ba = 56

Ba-147 = 146.93399 u

Q = E = Δmc²

= (239.0521565 - (91.911030 + 146.93399 + 3.02599503) * c²
= - 2.81886 * 931.5020
= -2625.77
= -2625.8 MeV (correct to 5 significant figures)

I think I've calculated it correctly now. I'm not sure about the negative/positive value. My notes say that 'if Q is negative, it represents the amount of energy which must be provided in order for the reaction to occur.'

 

[SteamKing]

Pu-239 ---> Sr-92 + Ba-147 + 3n

Z
Pu = 94
Sr = 38
Ba = 56

Q = E = Δmc²

= (239.0521565 - (91.911030 + 146.93399 + 3.02599503) * c²
= - 2.81886 * 931.5020
= -2625.77
= -2625.8 MeV (correct to 5 significant figures)

I think I've calculated it correctly now. I'm not sure about the negative/positive value. My notes say that 'if Q is negative, it represents the amount of energy which must be provided in order for the reaction to occur.'

Did you add up the atomic masses on both sides of the fission equation?

 

[atomicpedals]

Closer. However, you appear to have created mass with your fission products.

 

[says]

Left hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150
Right hand side = 239.0521565 + gamma

Q = E = Δmc²
= (m_initial - m_final) * c²

(239.0521565 + gamma)*c² ---> 241.8710150*c²

222,677.5619 - 225,303.3342
= - 2625.77 MeV
= - 2625.8 MeV (correct to 5 significant figures)

 

[SteamKing]

Left hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150
Right hand side = 239.0521565 + gamma

Q = E = Δmc²
= (m_initial - m_final) * c²

(239.0521565 + gamma)*c² ---> 241.8710150*c²

222,677.5619 - 225,303.3342
= - 2625.77 MeV
= - 2625.8 MeV (correct to 5 significant figures)

You only started with on atom of Pu-239, on the left hand side. How did you get "Left hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150"?

 

[says]

Sorry, meant to be

right hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150
left hand side = 239.0521565 + gamma

 

[SteamKing]

Your Q value is still about an order of magnitude too large.

 

[says]

Your Q value is still about an order of magnitude too large.

How could this be? I've used the correct units

Left hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150
Right hand side = 239.0521565 + gamma

Q = E = Δmc²
= (m_initial - m_final) * c²

(239.0521565 + gamma)*c² ---> 241.8710150*c²

222,677.5619 - 225,303.3342
= - 2625.77 MeV
= - 2625.8 MeV (correct to 5 significant figures)

 

[Astronuc]

Sorry, meant to be

right hand side = 91.911030 + 146.93399 + 3.02599503 = 241.8710150
left hand side = 239.0521565 + gamma

The atomic masses must be equal. 239 + 0 (gamma) = 92 + ? + 3.

 

[SteamKing]

How could this be? I've used the correct units

Generally, fission reactions are exothermic, greatly so. Why else would you want to build a nuclear plant, if you had to keep shoving all this energy into it to split atoms?

Also, fission reactions involving uranium and plutonium evolve pretty much the same energy per atom:

http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_7/4_7_1.html

 

[says]

The atomic masses must be equal. 239 + 0 (gamma) = 92 + ? + 3.

I assumed because this is a question about mass-energy equivalence, q-value / decay energy that the difference between products and reactants is what is needed to be input for the reaction to occur.

m_f*c² does not always equal m_i*c² because of the different binding energies of the product nuclei.

If Q = ∑m_fc² - m_ic²

(239.0521565)*c2 - 241.8710150*c2

-2.8188585 * 931.5020
= -2625.77 MeV

 

[atomicpedals]

You will have a mass difference, the sum of the masses of the products will be slightly smaller than that of the Pu. The "missing" mass being accounted for as energy.

The problem you have now is that your proposed fission product Ba results in the mass of the products being greater than the reactants, violating conservation of mass. Is there something on the periodic chart you could use that is slightly lighter than Ba?

 

[says]

I - 144 = 143.94961

So Pu-239 + gamma ---> Sr-92 + I-144 + 3n

Q = (239.0521565 + gamma) - (91.911030 + 143.94961 + (3*1.00866501)) * c2
= (239.0521565 + gamma) - 238.88663503
Q = 0.16552147 * c²
= 154.183580348 MeV

 

[atomicpedals]

Curious why you chose I-144 instead of Ba-144? Your Q value looks much better. On average fission of Pu-239 releases around 200MeV, so 154MeV is within what one might expect.

 

[says]

Actually,

Atomic Numbers
Pu = 94
Sr = 38
Ba = 56

Mass Numbers
Pu = 239
Sr = 92
Ba = 144
3 * neutrons = 3

Ba-144 = 143.922940

So Pu-239 + gamma ---> Sr-92 + Ba-144 + 3n

Q = (239.0521565 + gamma) - (91.911030 + 143.922940 + (3*1.00866501)) * c2
= (239.0521565 + gamma) - 238.85996503
Q = 0.19219147 * c2
= 179.026738688 MeV
=179.03 MeV (correct to 5 significant figures)

 

[says]

Curious why you chose I-144 instead of Ba-144? Your Q value looks much better. On average fission of Pu-239 releases around 200MeV, so 154MeV is within what one might expect.

I was getting confused with mass and atomic numbers. I think my post above this is correct though as I used Ba-144.

 

[atomicpedals]

Looks good to me! You've conserved Z number 38+56=94, and demonstrated energy/mass equivalence while still honoring mass conservation in the process.

 

[SteamKing]

If you look at the number of isotopes of barium and iodine, you'll see that each element has quite a number of these with a wide range of atomic weights.

http://en.wikipedia.org/wiki/Nuclear_fission#Fission_reactors

Fission reactions can result in varying numbers of neutrons produced to cause other fission reactions, and there is usually two different elements of varying atomic weight created when a uranium or plutonium nucleus is split.

This is why it is important to balance the number of nucleons on both sides of the reaction, just like it is important to balance the two sides of a chemical reaction.

 

[says]

That was confusing me for a while! I was looking at the atomic weights and not the atomic number or mass number. Nice to finally get to the right solution!