Question
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=r2Ce−kr. Suppose that k = 1fermi−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.
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=r2kq1q2
Complete answer:
Let us say there are two particles with charge q1 and q2 in space separated by a distance ‘r’. The electrostatic force between the two particles is given by,
F=r2kq1q2, k=4π∈∘1=9×109NmC−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=r2kq2
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=r2Ce−kr. Hence we can imply,
4π∈∘r2q2=r2Ce−kr⇒4π∈∘q2=Ce−kr∵q=1.6×10−19C, k=1fermi−1⇒9×109Nm2C−2(1.6×10−19)2C2=Ce−1(5)⇒C=23.04×10−29×e5Nm2=3.419∵e5=148.41,∴C=3.419×10−26Nm2
Therefore the value of ‘C’ is 3.419×10−26Nm2 .
Note:
∈∘ 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.