What is the ratio of the strength of the strong nuclear force to that of the electromagnetic force? based on this ratio, you might expect that the strong force dominates the nucleus, which is true for small nuclei. large nuclei, however, have sizes greater than the range of the strong nuclear force. at these sizes, the electromagnetic force begins to affect nuclear stability. these facts will be used to explain nuclear fusion and fission later in this text.

Vn/Ve = B

Explanation:

To establish this relationship we must examine the potentials that these forces create. The electrical potential is described by

         Ve = k q / r

The potential for strong nuclear force is

        Vn (r) = - gs / 4pir exp (-mrc / h)

Where gs is the stacking constant and r the distance between the nucleons,

We can compare these potentials where the force is derived from the relationship

        E = -dU / dr

        F = q E

The case of the electric potential varies with the inverse of the distance, in the distance between two nucleons is very strong

The case of strong nuclear potential has a competition between two factors:

- The inverse of the distance that is equal to the electromagnetic

- The negative exponential whose variation for small values of r is much greater than the inverse term of the distance and dominates the potential

Let's make an approximate relationship

Vn / Ve = [- gs / 4pir exp (-mrc / h)] / [k q / r]

Let's separate the constants and simplify the variables

Vn / Ve = [-gs / (4pi k q1)] exp (-A r)

A = mc / h

Vn/Ve = B

Vn = B Ve / exp (A r)

When r decreases the nuclear potential is the one that dominates.

double the distance = divide strength by 4