Question

What is the force of repulsion between two argon nuclei when separated by1nm (10-9 m)? The charge on an argon nucleus is +18e.

Answer 1

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Here's your Answer....

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Charge on an argon nuclei is, q = +18e = 18 × 1.6 × 10^-19 C

Distance between the two argon nuclei, r = 10-9 m

So, force of repulsion is, F = kq^2/r^2

=> F = (9 × 10^9)(18 × 1.6 × 10^-19)2/(10^-9)2

=> F = 2.592 × 10^-9 N
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Answer 2

Given:

The distance between the two argon molecules = 1 nm = $10^{-9}$m

The charge in one argon nucleus = +18e

The charge on one 1 electron = 1.6 × $10^{-19}$ Coulombs

To Find:

The force of repulsion between two argon atoms.

Solution:

The formula of the electrostatic force between two charged particles = $\frac{kq_{1}q_{2} }{r^{2} }$

where k = Electrostatic constant

q1 and q2 = charged particles

r = distance between two charged particles

Putting the values in the above formula

We get, F = $\frac{9*10^{9} *(18*1.6*10^{-19})^{2} }{10^{-18} }$ = 7.5 × $10^{-9}$ Newtons

Hence, the force of repulsion between the two argon nuclei is 7.5× $10^{-9}$ Newtons.