Final answer:
The Coulomb repulsion energy between two 16O nuclei can be calculated using Coulomb's Law by determining the charge of each nucleus and the distance between their centers, then converting the result from joules to MeV.
Step-by-step explanation:
To compute the Coulomb repulsion energy between two 16O nuclei that just touch at their surfaces, we use Coulomb's Law. First, we need to determine the charge of each nucleus, which is done by multiplying the atomic number of oxygen (8) by the elementary charge (e = 1.602 x 10-19 C). Since both nuclei have the same charge, the situation involves repulsive forces. The distance between the centers of the nuclei is twice the radius of one nucleus.
The formula for Coulomb's repulsion energy (U) is U = (k0 * q1 * q2) / r, where k0 is Coulomb's constant (8.987 x 109 Nm2C-2), q1 and q2 are the charges of the two nuclei, and r is the distance between the centers of the two nuclei. Since we are dealing with identical nuclei, the expression simplifies to U = (k0 * q^2) / (2R0). Given R0 = 1.2 x 10-15 m, we can calculate the energy.
To convert the result to megaelectronvolts (MeV), we will use the fact that 1 eV = 1.602 x 10-19 J. Multiplying the final value in joules by 106 and dividing by the charge of one electron gives us the energy in MeV.