Question

The force of interaction between two atoms is given by $F=\alpha \beta \exp \left({\displaystyle \frac{-x^2}{\alpha kt}}\right)$; 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 $F=\alpha \beta \exp \left({\displaystyle \frac{-x^2}{\alpha kt}}\right)$. 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 $\left({\displaystyle \frac{-x^2}{\alpha kt}}\right)$ which should be dimensionless.
Dimensionless quantity refers to $M^0{L}^0{T}^0$. So, $\left[{\displaystyle \frac{-x^2}{\alpha kt}}\right]=M^0{L}^0{T}^0$
Substituting the dimensions of all the known parameters in the equation $\left[{\displaystyle \frac{-x^2}{\alpha kt}}\right]=M^0{L}^0{T}^0$ to determine the dimension of α.
$$\begin{gathered} \left[{\displaystyle \frac{-x^2}{\alpha kt}}\right]=M^0{L}^0{T}^0 \\ \Rightarrow {\displaystyle \frac{L^2}{\alpha \times M{L}^2{T}^{-3}\times T}}=M^0{L}^0{T}^0 \\ \Rightarrow \left[\alpha \right]={\displaystyle \frac{L^2}{M{L}^2{T}^{-3}\times T}} \\ \Rightarrow \left[\alpha \right]=M^{-1}{L}^0{T}^2 \end{gathered}$$

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:
$$\begin{gathered} \left[F\right]=\left[\alpha \beta \exp \left({\displaystyle \frac{-x^2}{\alpha kt}}\right)\right] \\ =\left[\alpha \right]\times \left[\beta \right]\times \left[\exp \left({\displaystyle \frac{-x^2}{\alpha kt}}\right)\right] \end{gathered}$$
As, the dimensional unit of an exponential function is always 1 so, the above equation can be written as:
$$\begin{gathered} \left[F\right]=\left[\alpha \right]\times \left[\beta \right]\times \left[\exp \left({\displaystyle \frac{-x^2}{\alpha kt}}\right)\right] \\ =\left[\alpha \right]\times \left[\beta \right]----(i) \end{gathered}$$
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 $\beta$.
$$\begin{gathered} \left[F\right]=\left[\alpha \right]\times \left[\beta \right] \\ ⟹ML{T}^{-2}=M^{-1}{L}^0{T}^2\times \left[\beta \right] \\ ⟹\left[\beta \right]={\displaystyle \frac{ML{T}^{-2}}{M^{-1}{L}^0{T}^2}} \\ =M^{2}L{T}^{-4} \end{gathered}$$
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.