Question

It is common to illustrate the density of nuclear matter by discussing the mass of a teaspoon of nuclear material. What do nuclear densities imply for the mass of a teaspoon (5 cm³) of nuclear material? You may wish to use the relationship r = (1.2 fm)A^1/3 together with the mass of a proton, 1.67 × 10^-27 kg, to estimate.  

3.8 × 10¹¹ kg

1.53 × 10¹² kg

1.15 × 10¹² kg

9.6 × 10¹¹kg

Answer 1

Let the atomic number of the nucleus is A. And let the mass of a proton is m, then the total mass of the nucleus is M = mA.
And the radius of the nucleus is r, and so, the density of the nucleus is
   
And as given
   
where, r_0 = 1.2 fm, so, we have

And so, putting the values,
   ,
And so, we have
   
  
And given the volume of the spoon,
  
And so, the mass of the spoon assuming the nuclear matter is
   
  

So, 3rd option is correct.