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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 two protons situated at largest separation. Why do these protons not fly apart under this repulsion ?

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 two protons situated at largest separation. Why do these protons not fly apart under this repulsion ?

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1 Answers

Arun
25750 Points
5 years ago
Charge on each proton , Q = 1.6 × 10?¹?C 
separation between two protons , r = 6.9 fermi = 6.9 × 10?¹8 m 
Use formula , 
Here, Q1 = Q2 = Q = 1.6 × 10?¹?C 
so, F = 9 × 10? × (1.6 × 10?¹?)²/(6.9 × 10?¹5)² [ ? K = 9 × 10? Nm²/C² ]
= 9 × 2.56 × 10?³8?³°?/(6.9)² 
= 9 × 25.6/6.9 × 6.9 N 
= 4.84N
Hence, repulsive force between two protons is 4.84N
 The electromagnetic force of repulsion between positive protons in the nucleus is overcome by the strong nuclear force between protons and neutrons. This force is larger than the previous one and hence holds the nucleus together.So, these protons not fly apart under this repulsion.

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