What is the formula for calculating reaction energy in an α decay?

[Stephen Bulking]

Homework Statement

In α decay of Ra(A=226,Z=88) (at rest initially) : Ra radiates 3.7×10^10 α nuclei.The kinetic energy of an α nucleon is 4.78 MeV and the ratio between the mass of α and one of the daughter nucleon is mα/mdaughter =0.018.
The reaction energy is:
a) 2.88×10^-2 J
b) 50×10^-2 J
c) 30×10^-3 J
d) 0.85 J

Relevant Equations

Conservation of momentum
p1 + p2 = p3 + p4
Reaction energy
W= (M.initial - M.final)x c^2
Relation between momentum and kinetic energy
p^2 = 2mK

I tried momentum conservation, which gives:
-pα = pdaughter
<=> 2mKα = 2mKdaugther (squaring two sides)
Using the given mass ratio, I found Kdaughter to be 0.0864MeV
Adding the two Kinetic energy of the product particles and converting it to Joules, I got A
But I don't understand why adding the two kinetic energy of the product particles would yield the reaction energy (or does it? I'm not even sure I'm doing right, I just happen to get one of the numbers in the answers). The reaction energy is given by a different formula I put under the conservation of momentum formula and I don't think it says "add the two Kinetic energy together".

 

[Abhishek11235]

We have (Taking c=1, and v and v' to be particle velocities) :
$E_{init}= M_{nucleus}$
And
$E_{final}= M_{\alpha}+ M_{daughter}+ M_{\alpha}v^2/2 + M_{daughter}v'^2/2$

By conservation of Energy:
$$E_{init}=E_{final}$$

Combine the above equations!

 

[kuruman]

But I don't understand why adding the two kinetic energy of the product particles would yield the reaction energy (or does it? I'm not even sure I'm doing right, I just happen to get one of the numbers in the answers). The reaction energy is given by a different formula I put under the conservation of momentum formula and I don't think it says "add the two Kinetic energy together".

The reaction (or decay) energy is the total energy released in the reaction (or decay). Both particles are moving after the decay but were not moving before the decay. Is the reaction energy the kinetic energy of both particles together or just one of them? You are supposed to figure out on your own whether you should add the two kinetic energies together without being told.

 

[Stephen Bulking]

We have (Taking c=1, and v and v' to be particle velocities) :
Einit=Mnucleus
And
Efinal=Mα+Mdaughter+Mαv2/2+Mdaughterv′2/2

By conservation of Energy:
Einit=Efinal

Combine the above equations!

Thank you for your quick response. But I still have uncertainties, would you please help me out by explaining the following:
1) You wrote Einit = Mnucleus. I understand that this is the rest energy, and is equal to 226MeV. Using the second equation and conservation of momentum like I originally did would give me Efinal of 230.86604 MeV and this would contradict the third equation. I am unsure whether M in your equations is the same atomic mass of particles (MeV/c^2), which would be reasonable unit-wise.
2) Are you implying that the reaction energy is the sum of the two energies, which I do not find sensible and there is no answer with that result, or that the reaction energy is the subtraction of E initial from E final which would make sense and there is in fact an answer with that number (A). But if so, your third equation on conservation of energy should be restated as Einit + Ereact = Efinal.

 

[Stephen Bulking]

The reaction (or decay) energy is the total energy released in the reaction (or decay). Both particles are moving after the decay but were not moving before the decay. Is the reaction energy the kinetic energy of both particles together or just one of them? You are supposed to figure out on your own whether you should add the two kinetic energies together without being told.

Thank you for your quick response. I believe the answer to "Is the reaction energy the kinetic energy of both particles together or just one of them?" is both of them. But that is not where I am having a hard time with. My trouble lies in the fact that this method while proved successful by the results it yield, does not agree with the formula for reaction energy W= (M.initial - M.final)x c^2 very well. I think I must have misunderstood this formula somehow, please most kindly point out where it is that I misunderstood. Thanks again.

 

[kuruman]

Thank you for your quick response. I believe the answer to "Is the reaction energy the kinetic energy of both particles together or just one of them?" is both of them. But that is not where I am having a hard time with. My trouble lies in the fact that this method while proved successful by the results it yield, does not agree with the formula for reaction energy W= (M.initial - M.final)x c^2 very well. I think I must have misunderstood this formula somehow, please most kindly point out where it is that I misunderstood. Thanks again.

Relativistically, the total energy is $$E=T+mc^2$$ where $T$ is the kinetic energy. The energy before the decay is $$E_{before}=M_{parent}c^2.$$ The energy after is $$E_{after}=T_{daughter}+m_{daughter}c^2+T_{\alpha}+m_{\alpha}c^2.$$ Conserve energy and solve for the sum of the kinetic energies.

 

[Stephen Bulking]

Relativistically, the total energy is $$E=T+mc^2$$ where $T$ is the kinetic energy. The energy before the decay is $$E_{before}=M_{parent}c^2.$$ The energy after is $$E_{after}=T_{daughter}+m_{daughter}c^2+T_{\alpha}+m_{\alpha}c^2.$$ Conserve energy and solve for the sum of the kinetic energies.

Ok, thanks for your time, I really appreciate it.