Question

The force of interaction between two atoms is given by $F = \alpha \beta \exp \left( {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right)$; where x is the distance, k is the Boltzmann constant and T is temperature and α and β are two constants. The dimension of β is
A. ${M^2}{L^2}{T^{ - 2}}$
B. ${M^2}L{T^{ - 4}}$
C. ${M^0}{L^2}{T^{ - 4}}$
D. $ML{T^{ - 2}}$

Answer 1

Hint: In this question, we need to determine the dimension of β such that the force of interaction between two atoms is given by $F = \alpha \beta \exp \left( {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right)$. For this, we will apply the dimensional formula in each of the parameters and evaluate the dimension of β.

Complete step by step answer:
‘x’ is the distance and so, the dimensional unit of x is L.
‘k’ is the Boltzmann constant whose dimension is given as $M{L^2}{T^{ - 3}}$.
‘t’ is the temperature and so, the dimensional unit of ‘t’ is T.
α and β are two constants.
The raised power of the exponential function should always be a constant value with dimensionless terms. So, here the terms that are raised to the power of the exponential function is $\left( {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right)$ which should be dimensionless.
Dimensionless quantity refers to ${M^0}{L^0}{T^0}$. So, $\left[ {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right] = {M^0}{L^0}{T^0}$
Substituting the dimensions of all the known parameters in the equation $\left[ {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right] = {M^0}{L^0}{T^0}$ to determine the dimension of α.
\[
  \left[ {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right] = {M^0}{L^0}{T^0} \\
   \Rightarrow \dfrac{{{L^2}}}{{\alpha \times M{L^2}{T^{ - 3}} \times T}} = {M^0}{L^0}{T^0} \\
   \Rightarrow \left[ \alpha \right] = \dfrac{{{L^2}}}{{M{L^2}{T^{ - 3}} \times T}} \\
   \Rightarrow \left[ \alpha \right] = {M^{ - 1}}{L^0}{T^2} \\
 \]

Hence, the dimensional unit of the constant $\alpha $ is \[{M^{ - 1}}{L^0}{T^2}\].
Now, from the given equation we can write the dimensional equation as:
$
  \left[ F \right] = \left[ {\alpha \beta \exp \left( {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right)} \right] \\
   = \left[ \alpha \right] \times \left[ \beta \right] \times \left[ {\exp \left( {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right)} \right] \\
 $
As, the dimensional unit of an exponential function is always 1 so, the above equation can be written as:
$
  \left[ F \right] = \left[ \alpha \right] \times \left[ \beta \right] \times \left[ {\exp \left( {\dfrac{{ - {x^2}}}{{\alpha kt}}} \right)} \right] \\
   = \left[ \alpha \right] \times \left[ \beta \right] - - - - (i) \\
 $
Force is the product of the mass and the acceleration whose dimensional formula is given as $\left[ F \right] = ML{T^{ - 2}}$. Also, we know the dimensional unit of the constant $\alpha $ is \[{M^{ - 1}}{L^0}{T^2}\]. So, substitute the dimension of $\alpha $ and F in the equation (i) to determine the dimensional unit of $\beta $.
$
  \left[ F \right] = \left[ \alpha \right] \times \left[ \beta \right] \\
\implies ML{T^{ - 2}} = {M^{ - 1}}{L^0}{T^2} \times \left[ \beta \right] \\
 \implies \left[ \beta \right] = \dfrac{{ML{T^{ - 2}}}}{{{M^{ - 1}}{L^0}{T^2}}} \\
   = {M^2}L{T^{ - 4}} \\
 $
Hence, the dimension of the constant $\beta $ is given as ${M^2}L{T^{ - 4}}$.

So, the correct answer is “Option B.

Note:
Dimensions are the physical unit of the parameter. There are seven pre-defined dimensions in mathematics, based on which all the other measuring unit’s dimensions are defined such as Mass, ampere, length, temperature, candela, mole and time.