Question

Calculate the coulombic energy difference between the Calcium-39 and Potassium-39 mirror pair. Also calculate the difference between the Nitrogen-13 and Carbon-13 pair. How do these compare to the energy obtained from mass balance principles?

Answer 1

In a nucleus having Z protons and mass number A, Z nucleons are protons, each of which carries a charge +e. Thus the total charge is Ze(positive). In mirror pairs,total no of nucleons A is equal ,however the no of proton and neutrons are just opposite .

coulombic energy difference between the Calcium-39 and Potassium-39

For a pair of mirror nuclei of radius R, charges Ze and (Z-1)e,the coulomb energy

difference is given by,

∆E=(3/5 e^2/R)(2Z+1)

R=radius of nuclei

R=Ro A^1/3 where Ro=1.2*10^-15 m

R=1.2*10^-15 m * (39)^1/3=15.6*10^-15 m

Ca-39 Z=20, A=39

K-39 Z=19,A=39

1 amu = 1.66-10^-27 kg m                             

∆E=(1/4 eo)(3/5 e^2/R)(2Z+1)       where 1/4 eo=coulomb factor=8.99*10^9 Nm2/C2

    =( 8.99*10^9 Nm2/C2 )3/5 (1.6*10^-19 C)^2/(15.6*-10^-15 m) (2*20 +1)

    =(8.99*10^9 Nm2/C2) (4.03*10^-23 C^2 /m)

=    36.22 *10^-14 J

For Nitrogen-13 and Carbon-13 pair.       

∆E=(1/4 eo)(3/5 e^2/R ) *(2Z+1)

Z=7(N-13)

A=13

R=1.2*10^-15 m * (13)^1/3=5.2*10^-15 m

∆E=(1/4 eo) * 3/5 (1.6*10^-19 C)^2/(5.2*10^-15 m )(2*7+1)=( 8.99*10^9 Nm2/C2 )*(4.43*10^-23 C^2 m)=39.83*10^-14 J

2) When protons and neutrons form the nucleus, there is some mass defect in the actual mass of the nucleus.This loss of mass is released in the form of energy during the formation of nuclei, as per Einstein’s mass energy relation E=mc2.it is the binding energy of the nucleons of the nucleus.

If we consider the principle of symmetry according to which the nuclear force between a pair of protons and a pair of neutrons is same, then the mirror nuclei, (with different numbers of protons and neutrons ) should have the same binding forces.But the difference exists as the mirror nuclei with more protons has high charge density and so high coulomb self energy.