Question

At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons and a radius of 5.9 × 10−15 m. What is the magnitude of the repulsive force pushing these two spheres apart? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2

Answer 1

F = 3505.2 N

Step-by-step explanation:

Coulomb law states that the electrostatic force between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the two particles. The electrostatic force is given by:

`F=K(q_(1)q_(2) )/(r^(2) )`

Given:

coulomb constant(K) = 8.98755 × 10⁹ Nm²/C²

Charge of each particle(q₁ = q₂ = 46e)

e is the charge of each electron = 1.602 × 10⁻¹⁹ C

q₁ = q₂ = 46e = 46 × 1.602 × 10⁻¹⁹ = 73.692 × 10⁻¹⁹ C

The distance between the two particles(r) = r₁ + r₂

r₁ and r₂ are the radius of the protons

Therefore r = r₁ and r₂ = (5.9 × 10⁻¹⁵)m + (5.9 × 10⁻¹⁵)m = (11.8 × 10⁻¹⁵)m

Substituting values:

`F=8.98755*10^(9)( (73.692*10^(-19) *73.692*10^(-19) )/((11.8*10^(-15) )^(2) ))`

F = 3505.2 N

The magnitude of the repulsive force pushing these two spheres apart is 3505.2 N