The most common kind of iron nucleus has a mass number of `56`. Find the approximate density of the nucleus.

Explanation

We use two key ideas. The radius and mass of a nucleus depend on the mass number A and density is mass divided by volume. We use Eq.(i) to determine the radius of the nucleus. The mass of the nucleus in aatomic mass units is approximately equal to the mass number. The radius is
`R = R_(0) A^(1//3)`
` =(1.2 xx 1.0^(-15) m) (56)^(1//3)`
`=4.6 xx 10^(-15) m =4.6 fm`
Since `A =56`, the mass of the nucleus is approximately 56 u, or ` m~~(56)(1.66 xx 10^(-27) kg) =9.3 xx 10^(-26) kg`. The volume is
`nu =(4)/(3) pi R^(3) =(4)/(3) xx 3.14 xx(4.6 xx10^(-15))^(3)`
`~~ 4.1 xx 10^(-13) m^(3)`
And the density `rho` is approxiimately
`rho=(m)/(V)~~(9.3 xx 10^(-26) kg)/(4.1 xx 10^(-43) m^(3)) =2.3 xx 10^(27) kg m^(-3)`
The density of solid iron `700 kg m^(3)`, So we see that the iron nucleus is more than `10^(13)` times as dense as the bulk material. Densities of this magnitude are also found in neutron stars, which are similar to gigantic nuclei made almost entirely of neutrons. A `1 cm` cube of material with this density would have a mass of `2.3 xx 10^(11) kg` or `230` million metric tons.