a. Find an approximate expression for the mass of a nucleus of mass number `A`. b. Find an expression for the volume of this nucleus in terms of the mass number. c. Find a numerical value for its density.

Explanation

(a) The mass of the proton is approximately equal to that of neutron. Thus, if the mass of one of these particles is m, the mass of the nucleus is approximately Am.
(b) Assuming the nucleus is spherical and using Eq. (i), we find that the volume is
`V =(4)/(3)pi r^(3) =(4)/(3) pir_(0)^(3)A`
( C) The nuclear density can be found as follows:
`rho_(n) = (mass)/(volume) =(Am)/((4)/(3) pi r_(0)^(3) A) =(3m)/(4 pi r_(0)^(3))`
The fact that `A` cancels out of our expression for nuclear density proves our statemnet above that all nuclei have roughly the same density. Taking `r_(0) =1.2 xx 10^(-15)m` and `m =1.67 xx 10^(-27) kg`, we find that
`rho_(n) =(3(1.67 xx 10^(-27) kg))/(4 pi (1.2 xx 10^(-15) m)^(3))=2.3xx10^(17) kg m^(-3)`
Note that the nuclear density is ablout `2.3 xx 10^(14)` times greater than the density of water `(1 xx 10^(3) kg m^(-3))`.