KE of alpha particle using integer values of nuclear masses

[moenste]

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, E_α, of the alpha particle.
(b) (iii) Calculate E_α, 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?

 

[mfb]

(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.

 

[moenste]

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?

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...

 

[mfb]

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.

Decay makes the nucleus more stable. Aside from that not sure what to add...

Which conservation laws could be relevant in the decay?

 

[moenste]

No, you have to find the right decay process. Co-60 does not decay to Co-59.

Maybe something like ⁶⁰Co → ⁵⁹Co + ¹₀n?

Which conservation laws could be relevant in the decay?

Conservation of energy?

 

[mfb]

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?

 

[moenste]

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?

 

[mfb]

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.

 

[moenste]

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?

 

[mfb]

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.

 

[moenste]

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: p_α = m_α v_α, p_nucleus = m_nucleus v_nucleus.

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

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, E_α = E / (1 + m_α / m_nucleus) = 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.

 

[mfb]

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

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, E_α = E / (1 + m_α / m_nucleus) = 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.

 

[moenste]

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.

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.

(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.

 

[mfb]

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.

 

[moenste]

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 v_alpha + m v_nucleus = 206 v_nucleus + 4 v_alpha
0 = 206 v_nucleus + 4 v_alpha
206 v_nucleus = - 4 v_alpha

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

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.

 

[mfb]

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.

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.

 

[moenste]

It is not unnecessary, it is the only way to derive the equation you found.

m_D v_D = m_a v_a
KE_D + KE_a = 5.404 MeV

KE_D = 1/2 m_D v_D²
v_D = m_a v_a / m_D
KE_D = 1/2 m_D m_a² v_a² / m_D²
KE_a = 1/2 m_a v_a²
KE_D = 1/2 m_a v_a² * m_a / m_D
KE_D = KE_a * m_a / m_D

KE_a * m_a / m_D + KE_a = 5.404 MeV
Take KE_a out of the brackets: KE_a (m_a / m_D + 1) = 5.404 MeV
KE_a = 5.404 MeV / (m_a / m_D + 1)

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?

 

[mfb]

Looks good with the energy calculation.

(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.

 

[moenste]

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.

(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?

 

[mfb]

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.

 

[moenste]

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?

 

[mfb]

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?

 

[moenste]

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.

 

[mfb]

That is the right decay.