Order of magnitude of density of uranium nucleus is , [m = 1.67 xx 10^(-27 kg]`

Options

A. `10^(20) kg//m^(3)`
B. `10^(17) kg//m^(3)`
C. `10^(14) kg//m^(3)`
D. `10^(11) kg//m^(3)`

Answer

B

Explanation

Radius of a nucleus is given by
`R=R_(0)A^(1//3)` (where `R_(0)=1.25xx10^(-15) m)`
`=1.25 A^(1//3)xx10^(-15) m`
Here, A is the mass number and mass of the uranium nucleus will be `m~~Am_(p)`
`m_(p)=` mass of proton `=A(1.67xx10^(-27) kg)`
`:.` Density `rho=("mass")/("Volume")=m/(4/3 pi R^(3))=(A(1.67xx10^(-27)kg))/(A(1.25xx10^(-15) m)^(3))`
`implies rho~~2.0xx10^(17) kg//m^(3)`