The nuclear radius of `._(8)O^(16)` is `3 xx10^(-15) m`. If an atomic mass unit is `1.67 xx 10^(-27) kg`, then the nuclear density is approximately.

Answer

`2.359xx10^(17)kg//m^(3)`

Explanation

Here, `R=3xx10^(-15)m, rho=?`
Mass, `m=16a.m.u =16xx1.66xx10^(-27)kg`
`rho=m/V=m/(4/3piR^(3))=(3m)/(4piR^(3))`
`=(3xx16xx1.66xx10^(-27))/(4xx3.14(3xx10^(-15))^(3))=2.359xx10^(17)kg//m^(3)`