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

Hint: 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.