Question

Suppose the magnitude of Nuclear force between two protons varies with the distance between them as shown in figure. Estimate the ratio Nuclear force/Coulomb force" for

Answer 1

Answer: Here is your answer!

In the given figure force on the Y-axis is plotted on a logarithmic scale. Magnitude of Nuclear forces from the given figure by logarithmic interpolations are, For x=8 fm, => 0.18 N For x=4 fm, => 1.00 N For x=2 fm, => 17.7 N For x=1 fm, => 177 N Now Coulomb force is given by, F=q²/4π ε0 r² = 9x109 .e²/r² = 9x109 .(1.6x10-19 )²/r². (10-15)2 = 144/r² N (Here r is in fm) So this force, for x=8 fm, => 2.25 N for x=4 fm, => 9.00 N for x=2 fm, => 36.00 N for x=1 fm, => 144.0 N So the ratio "Nuclear force/Coulomb force" (a) for x=8 fm is 0.18 N/2.25 N = 0.08 (b) for x=4 fm is 1.0 N/9.0 N = 0.11 (c) for x=2 fm is 17.7 N/36.0 N = 0.491 (d) for x=1 fm is 177 N/144 N = 1.23 It is clear that in the range of r< 2 fm Nuclear forces are quite comparable to Coulomb forces.

Explanation: Hope it will help you! And I am very glad to help you! Thank You!

Answer 2

Coulomb forces are equal to the nuclear forces.

Explanation:

On the Y-axis of the given figure, the force is shown on the Y-axis and it is plotted on a logarithmic scale. From the given figure, the Nuclear forces has the magnitude by logarithmic interpolations are,  

$\text { For } x=4 \mathrm{fm},=>1.00 \mathrm{N}$

$\text { For } x=8 \mathrm{fm},=>0.18 \mathrm{N}$

$\text { For } x=1 \mathrm{fm}, \Rightarrow 177 \mathrm{N}$

$\text { For } x=2 \mathrm{fm},=>17.7 \mathrm{w} \mathrm{N}$

We need to now calculate the coulomb force between the distance of two protons, which is 8 fm.

So, the force is for $x=8 \mathrm{fm}, 2.25 \mathrm{N}$. $For x=2 fm$ then $36.00 \mathrm{N}$. For $x=4 \mathrm{fm}$, then, $9.00 \mathrm{N}$. And, for$x=1 \mathrm{fm}$, then $144.0$$N$.

It can be concluded that the Columb forces are equal to the nuclear forces.