KE of alpha particle using integer values of nuclear masses

In summary: MeV.In summary, the isotope 60Co is expected to undergo neutron emission as a form of radioactive decay due to the difference in nucleon number. The energy released in the disintegration of 21084Po to 20682Pb is 5.40(4) MeV, while the energy of the alpha particle does not all appear as kinetic energy due to momentum conservation. By using the formula Eα = E / (1 + mα / mnucleus), the calculated value for the kinetic energy of the alpha particle is 5.30(2) MeV.
  • #1
moenste
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Homework Statement


(a) Cobalt has only one stable isotope, 59Co. What form of radioactive decay would you expect the isotope 60Co to undergo? Give a reason for your answer.

(b) The radioactive nuclei 21084Po emit alpha particles of a single energy, the product nuclei being 20682Pb.
(b) (i) Using the data below, calculate the energy, in MeV, released in each disintegration.
(b) (ii) Explain why this energy does not all appear as kinetic energy, Eα, of the alpha particle.
(b) (iii) Calculate Eα, taking integer values of the nuclear masses.

Nucleus, mass (u): 21084Po, 209.936 730; 20682Pb, 205.929 421; α-particle, 4.001 504.

(1 atomic mass unit, u = 931 MeV.)

Answers: (b) (i) 5.40(4) MeV, (iii) 5.30(2) MeV.

2. The attempt at a solution
(a) No idea. It can't be α, β, since the difference in the nucleon number is 1: 60Co → 59Co.

(b) (i) (209.936 730 - 205.929 421 - 4.001 504) * 931 = 5.404455 MeV.

(b) (ii) No idea. KE like 1 / 2 m v2? Or E = m c2? Maybe because we don't have mass in kg in this situation?

(b) (iii) Also no idea. Integer values like 209 - 205 - 4? Without digits?

Any help please?
 
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  • #2
moenste said:
(a) No idea. It can't be α, β, since the difference in the nucleon number is 1: 60Co → 59Co.
That is not the decay process that happens. There are nuclei which do that (neutron emission), but those are rare.
(b) (i) (209.936 730 - 205.929 421 - 4.001 504) * 931 = 5.404455 MeV.
Missing units, but the answer looks good.
(b) (ii) No idea. KE like 1 / 2 m v2? Or E = m c2? Maybe because we don't have mass in kg in this situation?
Wrong direction. Assume that the initial nucleus is at rest. What happens after the decay? The alpha particle flies away - what about the remaining nucleus?
(b) (iii) Also no idea. Integer values like 209 - 205 - 4? Without digits?
You'll need (ii) to start here.
 
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  • #3
mfb said:
That is not the decay process that happens. There are nuclei which do that (neutron emission), but those are rare.
So that is the answer for (a)? Form of radioactive decay -- neutron emission?

mfb said:
Wrong direction. Assume that the initial nucleus is at rest. What happens after the decay? The alpha particle flies away - what about the remaining nucleus?
Decay makes the nucleus more stable. Aside from that not sure what to add...
 
  • #4
moenste said:
So that is the answer for (a)? Form of radioactive decay -- neutron emission?
No, you have to find the right decay process. Co-60 does not decay to Co-59.
moenste said:
Decay makes the nucleus more stable. Aside from that not sure what to add...
Which conservation laws could be relevant in the decay?
 
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  • #5
mfb said:
No, you have to find the right decay process. Co-60 does not decay to Co-59.
Maybe something like 60Co → 59Co + 10n?

mfb said:
Which conservation laws could be relevant in the decay?
Conservation of energy?
 
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  • #6
moenste said:
Maybe something like 60Co → 59C + 10n?
Assuming "C" is a typo (carbon??): No, that is the process where I said already that it does not happen.
Conservation of energy?
That is relevant, but not the point. Can the alpha particle just start moving in some direction while everything else stays at rest all the time?
 
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  • #7
mfb said:
Assuming "C" is a typo (carbon??): No, that is the process where I said already that it does not happen.
That is relevant, but not the point. Can the alpha particle just start moving in some direction while everything else stays at rest all the time?
I have no idea.

I searched for "kinetic energy alpha particle" and got tp this topic, but I don't have the mass and velocity. Found pictures like this and tried to calculate the answer using 209.936 730 - 205 - 4 and that didn't help. Integer values says that the number should have no digits (at least I understood the definition so). But 209 - 205 - 4 = 0 and doesn't help much.

No idea what to do and how to answer your question : (. What shall read on this?
 
  • #8
What about momentum conservation?

Your initial nucleus is at rest, if the alpha particle flies away in one direction, the remaining nucleus has to fly away in the opposite direction.

Don't start with (iii) before you got (ii) right, that won't work.
 
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  • #9
mfb said:
What about momentum conservation?

Your initial nucleus is at rest, if the alpha particle flies away in one direction, the remaining nucleus has to fly away in the opposite direction.

Don't start with (iii) before you got (ii) right, that won't work.
I understand the concept of momentum conservation. So p = m v initial and then find the velocities of the alpha particle and the nucleus?
 
  • #10
The initial nucleus is at rest, the total momentum is zero. After the decay the alpha particle moves in one direction, the remaining nucleus moves in the opposite direction. What does that mean for the energy? The alpha particle has some kinetic energy, what about the nucleus?

For (iii): you know the total energy released, you know the total momentum, and you know the particle masses, that should allow to find the energy of the alpha particle.
 
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  • #11
mfb said:
The initial nucleus is at rest, the total momentum is zero. After the decay the alpha particle moves in one direction, the remaining nucleus moves in the opposite direction. What does that mean for the energy? The alpha particle has some kinetic energy, what about the nucleus?

For (iii): you know the total energy released, you know the total momentum, and you know the particle masses, that should allow to find the energy of the alpha particle.
Time 0: p = m v = m * 0 = 0.
Time 1: pα = mα vα, pnucleus = mnucleus vnucleus.

They both have some kinetic energy and that is why the energy in (i) does not all appear as kinetic energy Eα?

Total energy released = 5.404455 MeV.
Total momentum = 0.
Particle masses = 205.929 421 u and 4.001 594 u.

KE = p2 / 2 m? But p = 0.

Update
I found a formula here, Eα = E / (1 + mα / mnucleus) = 5.404 455 / (1 + 4.001 504 / 205.929 421) = 5.301 404 MeV. Is this correct? Thought It's not 5.30(2) as in the answer.
 
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  • #12
moenste said:
They both have some kinetic energy and that is why the energy in (i) does not all appear as kinetic energy Eα?
Correct.

[quote[Particle masses = 205.929 421 u and 4.001 594 u.[/quote]You are supposed to use 206 and 4 here.
KE = p2 / 2 m? But p = 0.
##p \neq 0## for both the alpha particle and the produced nucleus.

Update
I found a formula here, Eα = E / (1 + mα / mnucleus) = 5.404 455 / (1 + 4.001 504 / 205.929 421) = 5.301 404 MeV. Is this correct? Thought It's not 5.30(2) as in the answer.
While that formula is right, I don't think you are supposed to use it without deriving it. It fits to the given answer, the agreement gets slightly better with the integer values for the masses.
 
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  • #13
mfb said:
Correct.
You are supposed to use 206 and 4 here.
##p \neq 0## for both the alpha particle and the produced nucleus.

While that formula is right, I don't think you are supposed to use it without deriving it. It fits to the given answer, the agreement gets slightly better with the integer values for the masses.
Yes, completely forgot about the integer values.

mfb said:
p \neq 0 for both the alpha particle and the produced nucleus.
But how to derive the momentum then (velocity is unknown)? Plus in KE = p2 / 2 m I think m only includes one mass of either the alpha particle or the produced nucleus.

moenste said:
(a) Cobalt has only one stable isotope, 59Co. What form of radioactive decay would you expect the isotope 60Co to undergo? Give a reason for your answer.
As I understand the original isotope Co-60 has one more neutron than Co-59.

6027Co → 5927Co + 10n.

I think it should be neutron emission.
 
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  • #14
moenste said:
But how to derive the momentum then (velocity is unknown)? Plus in KE = p2 / 2 m I think m only includes one mass of either the alpha particle or the produced nucleus.
You have two unknowns (the two velocities) and two equations - one for conserved momentum, one for the sum of energies of the two particles.

As I understand the original isotope Co-60 has one more neutron than Co-59.

6027Co → 5927Co + 10n.

I think it should be neutron emission.
No, it is not neutron emission. The daughter nucleus is NOT Co-59.
 
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  • #15
mfb said:
You have two unknowns (the two velocities) and two equations - one for conserved momentum, one for the sum of energies of the two particles.
p = m v = 0
p = m valpha + m vnucleus = 206 vnucleus + 4 valpha
0 = 206 vnucleus + 4 valpha
206 vnucleus = - 4 valpha

I think this is unnecessary complication. The momentum, the velocities. More chances to do a mistake somewhere.

mfb said:
No, it is not neutron emission. The daughter nucleus is NOT Co-59.
We have Co-60 that becomes Co-59. I did that and used neutron emission. I can also use proton emission, but in that case the proton number will decrease by one: 6027Co → 5926Co + 11proton emission.
 
  • #16
moenste said:
I think this is unnecessary complication. The momentum, the velocities. More chances to do a mistake somewhere.
It is not unnecessary, it is the only way to derive the equation you found.
moenste said:
We have Co-60 that becomes Co-59.
No it does not, and I wonder how often I have to repeat that.

There is no Cobalt with 26 protons, and proton emission is not the right decay either.
 
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  • #17
mfb said:
It is not unnecessary, it is the only way to derive the equation you found.
mD vD = ma va
KED + KEa = 5.404 MeV

KED = 1/2 mD vD2
vD = ma va / mD
KED = 1/2 mD ma2 va2 / mD2
KEa = 1/2 ma va2
KED = 1/2 ma va2 * ma / mD
KED = KEa * ma / mD

KEa * ma / mD + KEa = 5.404 MeV
Take KEa out of the brackets: KEa (ma / mD + 1) = 5.404 MeV
KEa = 5.404 MeV / (ma / mD + 1)

mfb said:
No it does not, and I wonder how often I have to repeat that.

There is no Cobalt with 26 protons, and proton emission is not the right decay either.
I can't find any other type of decay that fits this problem (the nucleon number decreases by one). Maybe not all of them are listed here?
 
  • #18
Looks good with the energy calculation.
moenste said:
(the nucleon number decreases by one)
It does not. I don't understand why you got stuck with that idea.

You can also look up the actual decay mode if you want. It is a very common one that you should be familiar with.
 
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  • #19
mfb said:
It does not. I don't understand why you got stuck with that idea.

You can also look up the actual decay mode if you want. It is a very common one that you should be familiar with.
moenste said:
(a) Cobalt has only one stable isotope, 59Co. What form of radioactive decay would you expect the isotope 60Co to undergo? Give a reason for your answer.
But it says that. We have Co-60 that decays to Co-59. Or I understand the text wrong?
 
  • #20
moenste said:
We have Co-60 that decays to Co-59
Where does it say that?
The problem statement says that Co-59 does not decay, and it says that Co-60 decays, but not to what. That's up to you to figure out.

It could also say that oxygen-16 is stable and Co-60 is not, that doesn't mean that Co-60 would decay to oxygen-16.
 
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  • #21
mfb said:
Where does it say that?
The problem statement says that Co-59 does not decay, and it says that Co-60 decays, but not to what. That's up to you to figure out.

It could also say that oxygen-16 is stable and Co-60 is not, that doesn't mean that Co-60 would decay to oxygen-16.
OK, now I understand what you mean.

But if we have Co-60 that decays, we can use any kind of decay (alpha, beta, gamma, etc.). What specific kind do they suggest?
 
  • #22
This is part of your homework task.

Which nuclei typically decay via alpha-decays? Does that sound plausible here?
Which nuclei typically decay via beta-decays? Does that sound plausible here?
Does a gamma decay change the isotope? Does that sound plausible here?

Did you finally look up which kind of decay happens?
 
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  • #23
mfb said:
This is part of your homework task.

Which nuclei typically decay via alpha-decays? Does that sound plausible here?
Which nuclei typically decay via beta-decays? Does that sound plausible here?
Does a gamma decay change the isotope? Does that sound plausible here?

Did you finally look up which kind of decay happens?
Beta decay is better for unstable nuclei + plus alpha-decay is usually used for "heavy" nuclei.

60Co → 60Ni + 0-1β.

I think this should be correct.
 
  • #24
That is the right decay.
 
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FAQ: KE of alpha particle using integer values of nuclear masses

1. What is the formula for calculating the kinetic energy of an alpha particle using integer values of nuclear masses?

The formula for calculating the kinetic energy of an alpha particle is KE = (m1 - m2)c^2, where m1 and m2 are the masses of the initial and final nuclei, respectively, and c is the speed of light. This formula can be used for calculating the kinetic energy of any particle, not just alpha particles.

2. How do we determine the masses of the initial and final nuclei when calculating the KE of an alpha particle?

The masses of the initial and final nuclei can be determined by looking at the atomic mass of each element. The atomic mass is the sum of the number of protons and neutrons in the nucleus. For example, an alpha particle is made up of two protons and two neutrons, so its mass would be equal to the mass of the initial and final nuclei in the KE formula.

3. Is the kinetic energy of an alpha particle affected by the distance between the nuclei?

No, the kinetic energy of an alpha particle is not affected by the distance between the nuclei. It is solely determined by the masses of the nuclei and the speed of the particle.

4. How does the speed of an alpha particle affect its kinetic energy?

The speed of an alpha particle is directly proportional to its kinetic energy. This means that as the speed increases, the kinetic energy also increases. The formula for calculating kinetic energy also includes the speed of light, which is a constant value of 299,792,458 meters per second.

5. How is the kinetic energy of an alpha particle related to its potential energy?

The kinetic energy of an alpha particle is related to its potential energy through the law of conservation of energy. As the alpha particle loses potential energy when moving towards a nucleus, it gains the same amount of kinetic energy. This means that the total energy of the system (alpha particle and nucleus) remains constant.

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