Calculation on spontaneous fission and effect from alpha decay

In summary, when calculating for spontaneous fission and the effect of alpha decay, one can use the initial power and divide it by the specific energy of Pu-238 to determine the mass required to generate the total thermal energy. This will give the activity, which is the product of the decay constant and the number of atoms present at a given time. To calculate the current effect, assuming the efficiency remains the same, one can use the initial power. Relevant equations for radioactive decay can be used to determine the rate of decay and the number of remaining undecayed atoms after a certain time.
  • #1
Luxdot
44
3
Summary:: Help needed on how to calculate on spontaneous fission and effect from alpha decay

Heat from alpha decay from Pu-238 is used to generate direct current. At the beginning (1977) it generated 470W, how large is the effect now? And if the efficiency between the electricity and heat transfer is 6%, how much did the fuel weight at the start?
Another question that I have is if a small fraction (7*10^-9 %) of the decay from U-235 occurs through spontaneous fission, how many spontaneous fissions occur per hour in 1 kg U-235. The half life time is 7,038*10^8 years.
 
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  • #2
One can take the initial power of 470 W and divide by the specific energy of Pu-238 to give the mass required to generate to total thermal energy. That will give the activity, A, and the activity is simply the product of the decay constant, λ, and the number of atoms present, N, at a given time.

hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/halfli2.html

From a mass of U-235, one can calculate the number of atoms present and multiply that number by the decay constant to get the activity, or decay rate, as explained in the Hyperphysics link.
 
  • #3
Welcome to PF. :smile:
Luxdot said:
Summary:: Help needed on how to calculate on spontaneous fission and effect from alpha decay

And if the efficiency between the electricity and heat transfer is 6%, how much did the fuel weight at the start?
This sounds like it could be a homework problem. Is this for schoolwork?
 
  • #4
Astronuc said:
One can take the initial power of 470 W and divide by the specific energy of Pu-238 to give the mass required to generate to total thermal energy. That will give the activity, A, and the activity is simply the product of the decay constant, λ, and the number of atoms present, N, at a given time.

hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/halfli2.html

From a mass of U-235, one can calculate the number of atoms present and multiply that number by the decay constant to get the activity, or decay rate, as explained in the Hyperphysics link.
Thank you! It’s a bit clearer now! However, I still don’t understand how I get the current effect….
 
  • #5
berkeman said:
Welcome to PF. :smile:

This sounds like it could be a homework problem. Is this for schoolwork?
Thanks! Kind of, I’m trying to learn a bit about nuclear power and found some old course material without solutions….. and I’m struggling a bit
 
  • #6
Okay, I'll move this to the schoolwork forums then. Even self-study problems go there. :smile:
 
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  • #7
Luxdot said:
Thank you! It’s a bit clearer now! However, I still don’t understand how I get the current effect….
If the 470 W is the electrical rather than the thermal power, then one should realize that electrical power (Pe) = thermal power (Pt) * efficiency, or conversely, Pt = Pe/efficiency.
 
  • #8
Astronuc said:
If the 470 W is the electrical rather than the thermal power, then one should realize that electrical power (Pe) = thermal power (Pt) * efficiency, or conversely, Pt = Pe/efficiency.
The 470 W comes from the radioisotope thermoelectric generator that uses the heat from the alpha decay of Pu-238. According to the task the first part is to calculate the current effect (470W was in 1977) and the second part is to calculate the weight of the fuel if the efficiency rate was 6%. So for the first part I shouldn’t use the mass of the fuel, right? Sorry if I am a bit confused, this is new to me….
 
  • #9
Luxdot said:
The 470 W comes from the radioisotope thermoelectric generator that uses the heat from the alpha decay of Pu-238. According to the task the first part is to calculate the current effect (470W was in 1977) and the second part is to calculate the weight of the fuel if the efficiency rate was 6%. So for the first part I shouldn’t use the mass of the fuel, right? Sorry if I am a bit confused, this is new to me….
Yes, for the first part you do not need the efficiency, you just have to assume it does not change.
Had you posted in the homework forum initially, you would have seen a template that invited you to state some Relevant Equations. Do you know or can you find any for radioactive decay? E.g. if you start with N atoms of a known half life, what is the rate of decay (atoms per unit time) and how many will remain undecayed after time t?
 
  • #10
haruspex said:
Yes, for the first part you do not need the efficiency, you just have to assume it does not change.
Had you posted in the homework forum initially, you would have seen a template that invited you to state some Relevant Equations. Do you know or can you find any for radioactive decay? E.g. if you start with N atoms of a known half life, what is the rate of decay (atoms per unit time) and how many will remain undecayed after time t?
I see! I know the formula for radioactive decay, but the initial number of atoms, N is not stated. Can this number be found or calculated for PU-235? The only known factors I have is the time, effect and that Pu-235 is used.
 
  • #11
Luxdot said:
I see! I know the formula for radioactive decay, but the initial number of atoms, N is not stated. Can this number be found or calculated for PU-235? The only known factors I have is the time, effect and that Pu-235 is used.
For the first part you do not need to know N. Just use it as an unknown as necessary and it will cancel out.
For the next part, you can calculate N from the amount of heat energy being released in total.
 
  • #12
haruspex said:
For the first part you do not need to know N. Just use it as an unknown as necessary and it will cancel out.
For the next part, you can calculate N from the amount of heat energy being released in total.
Okey, so N(t)=N(0)*e-lambda*t and lambda=ln(2)/T1/2. This gives me that lambda=0,0079. So N(t)=N(0)*e-0,0079*44. How do get rid of N(0) and I still don't see how I will get the current power?
 
  • #13
I solved the current power part! But I'm struggling a bit with the initial weight of the fuel. The way I've done it is by taking the electrical power of 470W and divided that by the efficiency to get the thermal power, this gives me 470/0,06=7834W. I then multiply this by the time (44 years in seconds) to get the Ws. 7834*(44*365*24*60*60) and then convert this to Joule, I get 1,087*1013 J. I then use E=mc2 and solve it for m. This gives me that m=1,087*1013/(3*108)2 = 1,2*10-4 kg. After some googling I found that the initial weight of the plutonium was around 2,4 kg, what have I done wrong?
 
  • #14
Luxdot said:
I then multiply this by the time (44 years in seconds)
But, as you calculated, the power has been diminishing those 44 years.
Anyway, you don't need to worry with how much total energy has been generated. If initially you got 7834W of thermal power, how many atoms were decaying per second? How many atoms does that imply in total at that time?
Luxdot said:
I then use E=mc2 and solve it for m.
And what will that m be the mass of?
 
  • #15
haruspex said:
But, as you calculated, the power has been diminishing those 44 years.
Anyway, you don't need to worry with how much total energy has been generated. If initially you got 7834W of thermal power, how many atoms were decaying per second? How many atoms does that imply in total at that time?

And what will that m be the mass of?
That makes sense, my problem is how I go from knowing the thermal power to how many atoms are decaying per second. The m would be the mass of the plutonium?
 
  • #16
Luxdot said:
The m would be the mass of the plutonium?
No, it is whatever mass is completely converted to energy. What is the mass of one Pu-238 atom and what is the sum of masses of the decay products?
How much mass has disappeared?
 
  • #17
haruspex said:
No, it is whatever mass is completely converted to energy. What is the mass of one Pu-238 atom and what is the sum of masses of the decay products?
How much mass has disappeared?
I see! The mass of 1 Pu-238 atom is 238/6,022*1023 = 3,95*10-22. How do I fid the total weight of the initial amount of plutonium from this?
 
  • #18
Luxdot said:
I see! The mass of 1 Pu-238 atom is 238/6,022*1023 = 3,95*10-22. How do I fid the total weight of the initial amount of plutonium from this?
What units?
As I wrote, you also need the masses of all the decay products. One is an alpha particle. What else?
 
  • #19
haruspex said:
What units?
As I wrote, you also need the masses of all the decay products. One is an alpha particle. What else?
Grams? It is not stated what the decay products are... Is this something that is the same for decay with plutonium? In the task it is stated that it uses the heat from the alpha decay. The question is about the Voyager 1 and how much plutonium it had in the beginning.
 
  • #20
Luxdot said:
Grams? It is not stated what the decay products are... Is this something that is the same for decay with plutonium? In the task it is stated that it uses the heat from the alpha decay. The question is about the Voyager 1 and how much plutonium it had in the beginning.
Luxdot said:
I then use E=mc2 and solve it for m. This gives me that m=1,087*1013/(3*108)2 = 1,2*10-4 kg.
This is the mass loss, which is a very small fraction of the Pu-238 remaining.

Luxdot said:
Okey, so N(t)=N(0)*e-lambda*t and lambda=ln(2)/T1/2. This gives me that lambda=0,0079. So N(t)=N(0)*e-0,0079*44. How do get rid of N(0)
One does not get rid of N(0)! One is trying to solve for N(0) in the first part of the problem.

N(0) is related to the total mass of the Pu-238.

The thermal power at any given time is related to the decay activity of the active isotope, which in this case is Pu-238. The total thermal power (Pt (t)) = e * A(t), where e is the energy per decay, and A(t) is the activity or decay rate (decays/time). A(t) = λ * N(t), where N(t) is the number of atoms of radioisotope of interest.

Focus on the Pu-238, which decays to U-234 by alpha decay. The half-life of Pu-238 is 87.7 y. The half-life of U-234 is 2.455×10+5 y, so it does not contribute much energy (heat) to the system, one because there is very little, and secondly because it has a relatively long half-life.
 
  • #21
Astronuc said:
This is the mass loss, which is a very small fraction of the Pu-238 remaining.One does not get rid of N(0)! One is trying to solve for N(0) in the first part of the problem.

N(0) is related to the total mass of the Pu-238.

The thermal power at any given time is related to the decay activity of the active isotope, which in this case is Pu-238. The total thermal power (Pt (t)) = e * A(t), where e is the energy per decay, and A(t) is the activity or decay rate (decays/time). A(t) = λ * N(t), where N(t) is the number of atoms of radioisotope of interest.

Focus on the Pu-238, which decays to U-234 by alpha decay. The half-life of Pu-238 is 87.7 y. The half-life of U-234 is 2.455×10+5 y, so it does not contribute much energy (heat) to the system, one because there is very little, and secondly because it has a relatively long half-life.
I still don't quite get it... Could you please elaborate a bit?
 
  • #22
Luxdot said:
I still don't quite get it... Could you please elaborate a bit?
What don't you get? I provided sufficient detail to solve the problem, and even provided a link to the Hyperphysics discussing the topic of radioactive decay.

For the current problem, if one determines the total energy (power integrated over time) and then calculates the mass producing that energy, one only calculates the mass of material consumed, not the initial mass, or the mass remaining at a given time.

Edit/update: After giving this problem some more thought, I realized that the time frame is 44 years, which is approximately half of one half-life (87.7 years), which would make it easier to solve.
 
Last edited:
  • #23
Solved it! Could you please help me to understand how I calculate the number of spontaneous fissions that occur every hour in 1 kg of U-235, knowing that 7*10-9% of the decay from U-235 occurs spontaneously.
 
  • #24
Luxdot said:
Could you please help me to understand how I calculate the number of spontaneous fissions that occur every hour in 1 kg of U-235,
Go back to here. What is the definition of activity, i.e., other words, what is one measuring (or calculating) when one measures (calculates) activity?
Astronuc said:
A(t) = λ * N(t), where N(t) is the number of atoms of radioisotope of interest.
 

Related to Calculation on spontaneous fission and effect from alpha decay

1. What is spontaneous fission and how does it occur?

Spontaneous fission is a type of nuclear decay where a heavy, unstable nucleus splits into two smaller nuclei without an external trigger. This process is caused by the strong nuclear force becoming weaker as the nucleus gets larger, making it unstable and prone to splitting into more stable forms.

2. How is the rate of spontaneous fission determined?

The rate of spontaneous fission is determined by the half-life of the radioactive material. This is the amount of time it takes for half of the material to decay. The shorter the half-life, the faster the rate of spontaneous fission.

3. What is the effect of alpha decay on spontaneous fission?

Alpha decay, where an alpha particle (two protons and two neutrons) is emitted from the nucleus, can increase the rate of spontaneous fission. This is because the loss of an alpha particle can destabilize the nucleus, making it more likely to undergo spontaneous fission.

4. How does the energy released from spontaneous fission and alpha decay compare?

The energy released from spontaneous fission is typically much larger than that of alpha decay. This is because spontaneous fission involves the splitting of a heavy nucleus, while alpha decay involves the emission of a small particle.

5. What are the potential applications of studying spontaneous fission and alpha decay?

Studying spontaneous fission and alpha decay can provide valuable information about the stability and properties of different elements and isotopes. This can be useful in nuclear power and weapons research, as well as in understanding the origins of the universe and the formation of heavy elements.

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