
Suppose an attractive nuclear force acts between two protons which may be written as $F=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}}$. Suppose that k = 1$\text{ferm}{{\text{i}}^{-1}}$ 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.
Answer
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Hint: 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:
$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Complete answer:
Let us say there are two particles with charge ${{q}_{1}}$ and ${{q}_{2}}$ in space separated by a distance ‘r’. The electrostatic force between the two particles is given by,
$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}},\text{ }k=\dfrac{1}{4\pi {{\in }_{\circ }}}=9\times {{10}^{9}}Nm{{C}^{-2}}$
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,
$F=\dfrac{k{{q}^{2}}}{{{r}^{2}}}$
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 $F=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}}$. Hence we can imply,
$\begin{align}
& \dfrac{{{q}^{2}}}{4\pi {{\in }_{\circ }}{{r}^{2}}}=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}} \\
& \Rightarrow \dfrac{{{q}^{2}}}{4\pi {{\in }_{\circ }}}=C{{e}^{-kr}} \\
& \because q=1.6\times {{10}^{-19}}C,\text{ }k=1\text{ferm}{{\text{i}}^{-1}} \\
& \Rightarrow 9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}{{(1.6\times {{10}^{-19}})}^{2}}{{C}^{2}}=C{{e}^{-1(5)}} \\
& \Rightarrow C=23.04\times {{10}^{-29}}\times {{e}^{5}}N{{m}^{2}}=3.419 \\
& \because {{e}^{5}}=148.41, \\
& \therefore C=3.419\times {{10}^{-26}}N{{m}^{2}} \\
\end{align}$
Therefore the value of ‘C’ is $3.419\times {{10}^{-26}}N{{m}^{2}}$ .
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.
Formula used:
$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Complete answer:
Let us say there are two particles with charge ${{q}_{1}}$ and ${{q}_{2}}$ in space separated by a distance ‘r’. The electrostatic force between the two particles is given by,
$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}},\text{ }k=\dfrac{1}{4\pi {{\in }_{\circ }}}=9\times {{10}^{9}}Nm{{C}^{-2}}$
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,
$F=\dfrac{k{{q}^{2}}}{{{r}^{2}}}$
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 $F=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}}$. Hence we can imply,
$\begin{align}
& \dfrac{{{q}^{2}}}{4\pi {{\in }_{\circ }}{{r}^{2}}}=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}} \\
& \Rightarrow \dfrac{{{q}^{2}}}{4\pi {{\in }_{\circ }}}=C{{e}^{-kr}} \\
& \because q=1.6\times {{10}^{-19}}C,\text{ }k=1\text{ferm}{{\text{i}}^{-1}} \\
& \Rightarrow 9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}{{(1.6\times {{10}^{-19}})}^{2}}{{C}^{2}}=C{{e}^{-1(5)}} \\
& \Rightarrow C=23.04\times {{10}^{-29}}\times {{e}^{5}}N{{m}^{2}}=3.419 \\
& \because {{e}^{5}}=148.41, \\
& \therefore C=3.419\times {{10}^{-26}}N{{m}^{2}} \\
\end{align}$
Therefore the value of ‘C’ is $3.419\times {{10}^{-26}}N{{m}^{2}}$ .
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.
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