from the relation `R=R_0A^(1//3)`, where `R_0` is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A ).

Explanation

Density of nucleus `(rho)` is defined as the nuclear mass per unit volume.
i.e. `rho =("Mass of one nucleus")/("Volume of nucleus")`
`=("Mass of one nucleon" xx "number of nucleon")/4/3 piR^(3)`
`=(mA)/(4/3piR^(3))`
But `R =R_(0)=R_(0)A^(1//3)` (given)
`m=1.66 xx 10^(-27)`kg
`therefore rho =(1.66 xx 10^(-27)A)/(4/3pi(R_(0)A^(1//3))^(3))`
`=(4.98 xx 10^(-27))/(4piR_(0)^(3)) = (4.98 xx 10^(-27))/(4 xx 22/7 xx (1.1 xx 10^(-15))^(3))`
`=2.97 xx 10^(17) kg//m^(3)`
Thus the nuclear density is independent of its mass number. Therefore, all nuclei have the same density approximately.