For a nucleus which is nearly spherical in shape `r=r_0 A^(1/3)`, where r is the radius and A is the mass number and `r_0` is a constant of value `1.2xx10^(-15)` m.
If the mass of the neutrons and protons are equal and equal to `1.67xx10^(-27)` kg, prove that density of the nucleus is `2.3xx10^14` times the density of water.

Explanation

Mass of A number of neutrons and proton `~~` mass of nucleus (M) =`1.67xx10^(-27) xx A kg `
Again, volume of the nucleus ,
`V=4/3pir^3=4/3pir_0^3A`
`=4/3xx3.14xx(1.2 xx 10^(-15))^3 . A m^3`
`therefore ` Density of nucleus `=M/V=(1.67xx10^(-27)xxA)/(4/3xx3.14xx(1.2xx10^(-15))^3.A)`
`=2.3xx10^17kg . m^3`
`therefore "density of nucleus "/"density of water " = (2.3xx10^17)/(1000) =2.3xx10^14`