Question

Considering the shape of calcium nucleus like a sphere calculate the density of nucleus of calcium of mass number 40

Answer 1

Answer:

Radius of Nucleus:

$R=R_{0} A^{1/3}$

Here $R_{0} =1.2*10^{-15} m$

Then  put the value in above formula we get

$R=1.2*10^{-15} *(40)^{1/3}$

$=4.10*10^{-15} m$

Then the density of the Nucleus of calcium:

Volume of Nucleus=(4/3)π$R^{3}$

=4/3*3.14*$(4.10*10^{-15} m)^{3}$

=4/3*3.14*$68.92*10^{-45} m^{3}$

=$2.88*10^{-43} m^{3}$

Density of nucleus=Mass/volume

The mass of the nucleus is 40

Density$=3.15*10^{-26} /2.88*10^{-43}$

$=1.38*10^{17} kg/m^{3}$