(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.

Relax

Respuesta :

Answer:

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.