Question

Consider A Gold Nucleus To Be A Sphere Of Radius 6.9 Fermi In Wch Protons And Neutrons Are Distributed.Find The Force Of Repulsion Between The Protons Situated At The Largest Separation

Answer 1

Largest separation between two protons in the nucleus is diameter of nucleus .
As you can see attachment. Both the protons P are separated 2r distance . 2r is the maximum distance between them , if we take distance > 2r , it means protons are not included in nucleus.

So, maximum separation between protons = 2r = 2 × 6.9 = 13.8 fermi
d = 13.8 × 10⁻¹⁵ m = 1.38 × 10⁻¹⁴ m
Charge on each proton , q = 1.6 × 10⁻¹⁹

Now, F = Kq²/r² [ from Coulombs law ]
F = 9 × 10⁹ × (1.6 × 10⁻¹⁹)²/(1.38 × 10⁻¹⁴)²
= 9 × 2.56/1.38 × 1.38 × 10⁹⁻³⁸⁺²⁸
= 12.09 × 10⁻¹ N
= 1.209 N

Answer 2

Explanation:

here is the answer for you dear