Question

(a) Evaluate, in MeV, the energy of gravitational attraction for two spherically symmetrical protons with a center-to-center separation of 2 F.

(b) Do the same for the energy of Coulomb repulsion at that separation.
(c) Compare your results with the energy of nuclear attraction, which is about —10 MeV at that separation.

Answer 1

Step-by-step explanation:

a ) Gravitational potential energy = - GMm/ r , G is universal gravitational constant , M and m are masses at distance r .

Putting the given values

Potential energy = -6.67 x 10⁻¹¹ x (1.67 x 10⁻²⁷)² / 2 x 10⁻¹⁵

=- 18.60 x 10⁻⁵⁰ J

=- 18.60 X 10⁻⁵⁰ / (1.6 x 10⁻¹⁹ x 10⁶) MeV

= - 11.62 x 10⁻³⁷ MeV

b ) electric potential energy = k q₁ x q₂ / r , k is constant which is equal to 9 x 10⁹ , q₁ and q₂ are charges and r is distance between charges

putting the values for two proton

= 9 x 10⁹ x ( 1.6 x 10⁻¹⁹ )² / 2 x 10⁻¹⁵ J

= 11.52 x 10⁻¹⁴ J

= 11.52 x 10⁻¹⁴ / (1.6 X 10⁻¹⁹ X 10⁶) MeV

= + .72 MeV

c ) energy of nuclear attraction = - 10 MeV.

It is more than energy of nuclear repulsion by more than 10 times.