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

Answer 1

Answer:

the force of repulsion between two protons situated at largest separation

can't understand the question sorry

Answer 2

Explanation:

Step 1:

Nucleus diameter is the maximum difference between two protons in the nucleus.

All protons P are 2r distance apart. 2r is the maximum distance between them, meaning that protons are not included in the nucleus if we take distance > 2r.

Step 2:

So, maximum distance between protons = $2 r=2 \times 6.9=13.8$ fermi

$r=13.8 \times 10^{-15} \mathrm{m}=1.38 \times 10^{-14} \mathrm{m}$

proton charge , $q=1.6 \times 10^{-19}$

Step 3:

Electrostatic force by Coulomb's Law,          

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$  

$=9 \times 10^{9} \times \frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(1.38 \times 10^{-14}\right)^{2}}$

$=9 \times 10^{9} \times \frac{2.56 \times 10^{-28}}{1.90 \times 10^{-28}}$

$=9 \times 10^{9} \times 1.34 \times 10^{-10}$

$=1.21 \mathrm{N}$