Starting a nuclear reaction In Chemistry, an amount of energy has to be supplied for a reaction to occur. This energy, known as the "activation energy", breaks up the bonds between molecues in the substance. It is equivalent to the total bond energy of the reactants.
However, in high school I learnt that the energy required to start a nuclear reaction is the difference between the binding energy of the reactants and the binding energy of the products.
Why is it that the minimum required energy is not the binding energy of the reactants, similar to a chemical reaction?
 A: 
In Chemistry, an amount of energy has to be supplied for a reaction to occur. This energy, known as the "activation energy", breaks up the bonds between molecules in the substance. It is equivalent to the total bond energy of the reactants.

In chemistry activation energy is not the energy required to break a bond.  Instead, the activation energy is the difference in energy between a transition state and the reactants.
A product bond can begin to form before the reactant bond is completely broken.
A transition state may be stabilized by a catalyst, lowering the activation energy.  This is critical to all life, as many biochemical reactions would not proceed at necessary rates without enzyme catalysis.

I learnt that the energy required to start a nuclear reaction is the difference between the binding energy of the reactants and the binding energy of the products.

As for nuclear reactions, the activation energy is not the difference between the binding energy of the reactants and the binding energy of the products.  That would be the negative of the energy released by the reaction.  For activation energy of a nuclear reaction, the energy difference between a transition state and the reactants should be considered.
A: For nuclear reactions we commonly talk about the Q of the reaction. for the reaction A(a,b)B where "A" and "a" are reacts (A is generally the target, "a" is the projectile) and "b" and "B" are the products,
$Q = \left(m(\mathrm{A})+m(\mathrm{a})-m(\mathrm{B})-m(\mathrm{b})\right)c^2.$ Here, $m(A)$ is the ${nuclear}$ mass of A. If $ Q>0$, the reaction will release energy in the center of mass; if Q<0, energy must be supplied to the center of mass frame. This value, however, does not take into account the energy required to overcome any coulomb repulsion which might be involved, the answer to your comment is:
No, balancing the Q value is not sufficient to making the reaction happen. If the reaction involves bringing two positive nuclei (or particles) together, you must provide energy to overcome coulomb repulsion.
