Given the atomic mass of `""^(56)Fe` is `55.93 u`, find its nuclear density. `[1 u = 1.66 xx 10^(-27) kg, R_(0) = 1.2 fm]`

Explanation

Data: `A=56, m = 55.93 u= 55.93 xx 1.66 xx 10^(-27)` kg, `R_(0) = 1.2 xx 10^(-15)` m
Nuclear radius, `R=R_(0)A^(1//3)`
Nuclear volume, `V=4/3 piR^(3) = 4/3 piR_(0)^(3)A`
`therefore` The nuclear density of `""^(56)Fe`,
`rho = m/V= m/(4/3pi R_(0)^(3)A) = (3m)/(4pi R_(0)^(3)A)`
`=(3(55.93 xx 1.66 xx 10^(-27)))/(4(3.14)(1.2 xx 10^(-10))^(3)(56))= 2.29 xx 10^(17) kg//m^(3)`