Question

Suppose an attractive nuclear force acts between two protons which may be written as $F = C{e^{ - kr}}/{r^2}$.
(A) Write the dimensional formula and
(B) Appropriate SI units.

Answer 1

Hint substitute the dimensional formulas of unknown physical quantities to derive the dimensional formula for unknown physical quantity. Similar steps can be followed to find SI units of any unknown physical quantity.

Complete Step By Step Solution
Given that, an attractive nuclear force acts between two protons which may be written as
$F = \dfrac{{C{e^{ - kr}}}}{{{r^2}}}$
We need to find the dimensional formula and dimensional formula is an expression which shows how and which of the fundamental units are required to represent the unit of a physical quantity.
All measurable physical quantities can be expressed in terms of seven fundamental quantities which are as follows:
1. Mass
2. Length
3. Time
4. Electric current
5. Temperature
6. Luminous intensity and
7. Amount of substance (mole)
In mechanics we consider three basic fundamental physical quantities to derive the dimensional formula of any other physical quantity which are: Mass, Length and Time.
Now, in the above expression the only unknown quantity is C which is a constant, F is force and r is the internuclear distance between the protons. Therefore, we need to find a dimensional formula for C.

$$\Rightarrow F = \dfrac{{C{e^{ - kr}}}}{{{r^2}}} \\\ \Rightarrow \dfrac{{F \times {r^2}}}{{{e^{ - kr}}}} = C \\\$$

We know, Force = mass $\times$acceleration,
Acceleration = $\dfrac{{velocity}}{{time}}$ where, velocity =$\dfrac{{{\text{distance}}}}{{{\text{time}}}}$ .
So, dimensional formula for speed/ velocity is given by: [${{\text{M}}^0}{L^1}{T^{ - 1}}$] because speed has no mass quantity in its formula so mass dimensions were${{\text{M}}^0}$, distance is measurement of length then length dimensions are ${L^1}$ and T is for time which is in denominator so ${T^{ - 1}}$are time dimensions.
$\Rightarrow velocity = [{{\text{M}}^0}{L^1}{T^{ - 1}}]$
Similarly, we can find dimensional formula for acceleration,
$\Rightarrow$Acceleration =$\dfrac{{[{{\text{M}}^0}{L^1}{T^{ - 1}}]}}{T}$
$\Rightarrow Acceleration{\text{ }} = [{{\text{M}}^0}{L^1}{T^{ - 2}}]$
Thus, dimensional formula for force can be given by:
$\Rightarrow$Force= [${M^1}] \times [{M^0}{L^1}{T^{ - 2}}]$
$\Rightarrow$Force= $[{M^1}{L^1}{T^{ - 2}}]$ OR
$\Rightarrow$ Force= $[ML{T^{ - 2}}]$
For${r^2}$, dimensional formula is given by, $[{L^2}]$because r is the internuclear distance between the protons and distance is a measurement of length.
\Rightarrow $$$${r^2} = [{L^2}]
Now, ${e^{ - kr}}$is a dimensionless quantity
$\Rightarrow {e^{ - kr}} = [{M^0}{L^0}{T^0}]$
Putting all above derived dimensional formula in below given equation
$\Rightarrow \dfrac{{F \times {r^2}}}{{{e^{ - kr}}}} = C$ …………………………………..eq.1
$\Rightarrow C = \dfrac{{[ML{T^{ - 2}}] \times [{L^2}]}}{{[{M^0}{L^0}{T^0}]}}$
$\Rightarrow C = [M{L^3}{T^{ - 2}}]$
So, the dimensional formula for C is$[M{L^3}{T^{ - 2}}]$.
Similarly, we can find the SI unit for C by putting the SI unit of force (F) and${r^2}$in equation 1
SI unit of force is Newton and r (distance) is meter and for ${r^2}$is $mete{r^2}$
$\Rightarrow C = N \times {m^2}$

Hence, the SI unit of C is$N{m^2}$.

Note key points to remember during derivation of dimensional formula for any physical quantity.
1. For any physical quantity, firstly derive its relation with other simplest physical quantities which can easily be represented in terms of mass, length and time.
2. For ${\left( {length} \right)^2}$, dimensional formula is $[{L^2}]$,
Write inverse power of quantity in numerator, if any fundamental quantity is in denominator, similar as we write dimensional formula for acceleration and speed.
3. If in any relationship of derived quantity, there is no term of mass or time or length , then represent that fundamental quantity raised to power zero in dimensional formula