Assuming the radius of a nucleus to be equal to `R=1.3 A^(1//3)xx10^(-15)m`. Where `A` is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus. Take mass of one nucleon `=1.67xx10^(-27)kg`

Answer

`2xx10^(11)kg//cm^(3), 1xx10^(38)n ucl//cm^(3)`

Explanation

The radius of nucleus is `R=R_(0)A^(1//3)`
where `A=` mass number `R_(0)=1.3xx10^(-15)m=1.3 fm`
The volume of nuclues is `=(4)/(3)piR^(3)=(4)/(3)pi(R_(0)A^(1//3))=(4)/(3)piR_(0)^(3)A`
A`=` mass number `=` number of necleons
`:.` The number of nucleons per unit volume is `=(A)/(V)`
`(1)/((4)/(3)piR_(0)^(3))=1.09xx10^(38)"necleons"//"cc"`
Density is " "`rho=("mass")/("volume")`
`=1.09xx10^(38)xx"mass of nucleons per cc"`
`=1.82xx10^(11)kg//"cc"`