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

The order of magnitude of mass and volume of uranium nucleus will be `m = A (1.67 xx 10^(-27) kg )` ( A is atomic number)
`V = (4)/(3) pi r^(3) = (4)/(3) pi [(1.25 xx 10^(-15) m) A^(1//3)]^(3)`
`= (8.2 xx 10^(-45) m^(3)) A`
Hence , `rho = (m)/(V) = (A (1.67 xx 10^(-27) kg))/((8.2 xx 10^(-45) m^(3)) A)`
`=2.0 xx 10^(17) kg //m ^(3)`