Binding energy in nuclear fission

Question

As far as i know, in a nuclear reaction we "go" from a binding energy $B_1$ to a binding energy $B_2$ with $B_2$>$B_1$ because a bigger binding energy means more stability for the nucleus.

If we consider the nuclear fission $n+^{235}U=^{141}Ba+^{92}Kr+3n$ the binding energies $B_1$ and $B_2$ are the sum of all the binding energies of components before and after the reaction?

Answer 1

In nuclear reactions, the binding energy is only part of the equation. It's more important to actually investigate the Q of the reaction which the the mass-energy difference: $$Q=M_1c^2 - M_2c^2,$$ where $M_1$ is the sum of the nuclear masses of the reactants (such as $n$ and $~^{235}U$), and $M_2$ is the sum of the nuclear masses of the products (such as $~^{141}Ba$, etc.).

The binding energies appear when you break the complex nuclei down into their individual nucleon masses. Again $M$ is a nuclear mass, not an atomic mass: $$M(Z,A)=Zm_p + (A_Z)m_n - B(Z,A)/c^2,$$ where $m_p$ is the proton mass, $m_n$ is the neutron mass, and $B(Z,A)$ is the binding energy of the nucleus. If you do the accounting with atomic mass values, it's a little different; you must account for the atomic electrons by using the atomic mass of $^1H$ instead of $m_p$.

Now let's talk about $Q$.

• If $Q$ is positive, the reaction could happen spontaneously, based on energy considerations. That doesn't mean it will happen, because other things may prohibit it or make different reactions more favorable. It is energetically possible for $~^{208}Pb$ to undergo alpha-decay, but the estimated lifetime is close to the age of the universe: we don't easily, if ever, observe it. The closer $Q$ is to zero, the lower the probability of the reaction proceeding.
• If $Q$ is negative, the reaction will not happen spontaneously. One would have to add energy (kinetic, electromagnetic) to make the reaction proceed.

In the case of fission, the number of protons and the number of neutrons remain constant (no beta or alpha decay), so $$Q = B(^{141}Ba)+B(^{92}Kr)-B(^{235}U)$$

In the case of beta decays, the numbers of protons and neutrons change, and you get an electron mass in the mix, so you have to be careful.