The energie distribution in nuclear fission

Question

When a nuclear fission occurs, the released energy is split between the reaction products: $$K_1=\frac{m_2}{m_2+m_1}E,$$ with $E$ the released energy, $m_2$ the mass of one of the products and $m_1$ of the other.

When calculating the total energy $E_1 = K_1+m_1c^2$, and doing the same for the product of mass $m_2$, these two energies aren't equal.

The energy released in a fission reaction isn't split equally between the products.

Why is this?

Answer 1

The decay must satisfy both the conservation of energy and conservation of linear momentum. I will not use a relativistic treatment here. Assume that an amount of energy $E$ is liberated (converted from rest energy to kinetic energy) during the fission.

We can then write down both conservation equations,

CLM: $m_{1}v_{1} = m_{2}v_{2}$

COE: $\frac{1}{2}m_{1}{v_{1}}^{2} + \frac{1}{2}m_{2}{v_{2}}^{2} = E$

We make the substitution $v_{1} = \frac{m_{2}v_{2}}{m_{1}}$ into the energy equation:

$m_{2}{v_{2}}^{2}(\frac{m_{2}}{2m_{1}} + \frac{1}{2}) = E \implies {v_{2}}^{2} = \frac{2m_{1}E}{m_{2}(m_{1}+m_{2})}$

Finally we can write down the kinetic energy of the second particle:

$\frac{1}{2}m_{2}{v_{2}}^{2} = \frac{m_{1}}{m_{1}+m_{2}}E$

We chose to number the particles arbitrarily, so by a sort of symmetry we can write down the kinetic energy of the other particle as

$\frac{1}{2}m_{1}{v_{1}}^{2} = \frac{m_{2}}{m_{1}+m_{2}}E$

So as you can see, we have derived that the kinetic energies of both of the particles are different! This must be so in order to satisfy the conservation laws. If we add the respective rest energies onto these, we would also find that the total energies are different.