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

Charge on each proton , Q = 1.6 × 10⁻¹⁹C
separation between two protons , r = 6.9 fermi = 6.9 × 10⁻¹⁸ m

Use formula ,
$\bold{F = \frac{KQ_1Q_2}{r^2}}$
Here, Q₁ = Q₂ = Q = 1.6 × 10⁻¹⁹C
so, F = 9 × 10⁹ × (1.6 × 10⁻¹⁹)²/(6.9 × 10⁻¹⁵)² [ ∵ K = 9 × 10⁹ Nm²/C² ]
= 9 × 2.56 × 10⁽⁻³⁸⁺³⁰⁺⁹⁾/(6.9)²
= 9 × 25.6/6.9 × 6.9 N
= 4.84N

Hence, repulsive force between two protons is 4.84N

Answer 2

Explanation:

here is the answer for you dear