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Question: Suppose an attractive nuclear force acts between two protons which may be written as \(F=\dfrac{C{{e...

Suppose an attractive nuclear force acts between two protons which may be written as F=Cekrr2F=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}}. Suppose that k = 1fermi1\text{ferm}{{\text{i}}^{-1}} and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.

Explanation

Solution

The protons have a positive charge and hence they repel each other. In the question it is given that at a distance of 5 fermi the effective nuclear force becomes equal to the electrostatic force of repulsion. Hence we can obtain the value of ‘C’ by equating the two forces respectively.
Formula used:
F=kq1q2r2F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}

Complete answer:
Let us say there are two particles with charge q1{{q}_{1}} and q2{{q}_{2}} in space separated by a distance ‘r’. The electrostatic force between the two particles is given by,
F=kq1q2r2, k=14π=9×109NmC2F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}},\text{ }k=\dfrac{1}{4\pi {{\in }_{\circ }}}=9\times {{10}^{9}}Nm{{C}^{-2}}
If two protons are at a distance ‘r’ from each other, there will be an electrostatic repulsion between them. Hence the equation of force of repulsion is,
F=kq2r2F=\dfrac{k{{q}^{2}}}{{{r}^{2}}}
In the question is given that when the distance between the two is 5 fermi, the electrostatic force of repulsion is equal to an attractive nuclear force acts between two protons which may be written as F=Cekrr2F=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}}. Hence we can imply,
q24πr2=Cekrr2 q24π=Cekr q=1.6×1019C, k=1fermi1 9×109Nm2C2(1.6×1019)2C2=Ce1(5) C=23.04×1029×e5Nm2=3.419 e5=148.41, C=3.419×1026Nm2 \begin{aligned} & \dfrac{{{q}^{2}}}{4\pi {{\in }_{\circ }}{{r}^{2}}}=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}} \\\ & \Rightarrow \dfrac{{{q}^{2}}}{4\pi {{\in }_{\circ }}}=C{{e}^{-kr}} \\\ & \because q=1.6\times {{10}^{-19}}C,\text{ }k=1\text{ferm}{{\text{i}}^{-1}} \\\ & \Rightarrow 9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}{{(1.6\times {{10}^{-19}})}^{2}}{{C}^{2}}=C{{e}^{-1(5)}} \\\ & \Rightarrow C=23.04\times {{10}^{-29}}\times {{e}^{5}}N{{m}^{2}}=3.419 \\\ & \because {{e}^{5}}=148.41, \\\ & \therefore C=3.419\times {{10}^{-26}}N{{m}^{2}} \\\ \end{aligned}
Therefore the value of ‘C’ is 3.419×1026Nm23.419\times {{10}^{-26}}N{{m}^{2}} .

Note:
{{\in }_{\circ }} is the permittivity of free space. It is to be noted that C is not dimensionless. Therefore we can say that it can vary with conditions. It also has a similar nature as that of the permittivity of free space but does not involve the charge on the particles.