Question

Let us assume that an attractive nuclear force has been experienced between two protons which can be shown as $F=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}}$.
(a) Write down the dimensional formula of $C$ and the appropriate SI units.

Answer 1

As we all know, the exponential function will be a dimensionless quantity. In this equation, therefore we can find that ${{e}^{-kr}}$ will be a dimensionless quantity. Not only the dimension, but also the quantity will be unit less. This will be helpful in answering this question.

Complete answer:
Let us mention the equation given in the question which can be written as,
$F=\dfrac{C{{e}^{-kr}}}{{{r}^{2}}}$
Rearranging this equation can be shown as,
$F{{r}^{2}}=C{{e}^{-kr}}$
Rearranging this equation can be written as,
$C=\dfrac{F{{r}^{2}}}{{{e}^{-kr}}}$
As we all know, the exponential function will be a dimensionless quantity.
In this equation, therefore we can find that ${{e}^{-kr}}$ will be a dimensionless quantity.
The dimension of the force can be shown as,
$\left[ F \right]=\left[ ML{{T}^{-2}} \right]$
The dimension of the square of the radius can be shown like this,
$\left[ {{r}^{2}} \right]=\left[ {{L}^{2}} \right]$
Substituting the values in the equation can be written as,
$C=\dfrac{\left[ ML{{T}^{-2}} \right]\times \left[ {{L}^{2}} \right]}{1}=\left[ M{{L}^{3}}{{T}^{-2}} \right]$
Therefore the dimensional value of the constant has been found. The SI unit of $C$ can be found as,
The SI unit of force will be $N$ and that of the square of the radius can be written as ${{m}^{2}}$. The exponential number will be a constant. Substituting this in the equation of the constant can be written as,
$C=N{{m}^{2}}$
Therefore the answer for the question has been calculated.

Note:
The nuclear force can be a force which will be experienced between the protons and neutrons of an atom. The nuclear force will be the force which binds the protons and neutrons in a nucleus together. This force will be found in between the protons and protons, neutrons and protons or neutrons and neutrons.