Question

Consider a gold nucleus to be a sphere of radius 6.9 fermi in which protons and neutrons are distributed. find the force of repulsion between the 2 protons situated at largest separation

Answer 1

I'm not sure but according to me there is repulsion/ repulsive force between them. Because all protons that are enclosed in a surrounding i.e. called nucleus are in equilibrium state. Otherwise,as they has great magnitude of charge on themselves,if they repel then the nucleus does not exists.
hope u understand..... but please confirm it from others too.....okkkk.
sorry not to answer up to the Mark.

Answer 2

Solution :-

Here radius of the gold nucleus is of 6.9 fermi and two protons are at two ends of the sphere.

So here distance between the two charges (r) would be equal to the diameter of the sphere.

• Distance b/w charges (r) = 2(Radius)
• Distance b/w charges (r) = 2 × 6.9
• Distance b/w charges (r) = 13.8 fermi

As we know that 1 fermi = 10^-15 m

Therefore, distance between the charges would be "13.8 × 10^-15".

As we know that electrostatic force between two charges is calculated by the given formula :

• F = kq₁q₂ / r²

Here,

• k is dielectric constant
• q₁ is value of first charge
• q₂ is value of second charge
• r is distance between charges.

Remember that value of both the charges would be charge on a single proton i.e., 1.6 × 10^-19 C

Substituting the values :

>> F = 9 × 10^-9 × (1.6 × 10^-19) × (1.6 × 10^-19) / )13.8 × 10^-15)²

>> F = 9 × 10^-9 × 1.6 × 1.6 × 10^-38 / 190.44 × 10^-30

After solving these complex calculations we get :

>> F = 1.2