Question

Rutherford found the size of the nucleus to be about 10^(-15)m. This implied a huge density. What would this density be for gold?

a) 1.92 × 10^17 kg/m^3
b) 3.84 × 10^17 kg/m^3
c) 7.68 × 10^17 kg/m^3
d) 9.60 × 10^17 kg/m^3

Answer 1

The estimated nuclear density for gold is approximately 1.92 × 10^17 kg/m^3, which can be calculated from the fact that nuclear densities are around 2 × 10^14 times the density of water.

Step-by-step explanation:

Rutherford's discovery indicated that the nucleus of an atom is very small, about 10-15 meters in size, and contains nearly all of the atom's mass, leading to extremely high densities in nuclear matter. For gold, which has a nucleus roughly the same order of magnitude in size, we can estimate its density.

Considering that nuclear densities are about 2 × 1014 times greater than that of water, and water has a density of approximately 1 × 103 kg/m3, it follows that the density of a nucleus would be around 2 × 1017 kg/m3. This results in the density of gold's nucleus being roughly 1.92 × 1017 kg/m3, corresponding to answer (a).