An expression for a dimensionless quantity P is given by $P=\frac{\alpha }{\beta }{\log }_e\left(\frac{kt}{\beta x}\right);$ where α and β are constants, x is distance; k is the Boltzmann constant and t is the temperature. Then the dimensions of α will be :

CONCEPT:

• Boltzmann's constant is the proportionality constant that relates to the average relative kinetic energy of the particles in a gas with the thermodynamics temperature of the gas. Its value is written as;

           k = 1.380649 $\times$ 10^-23 m² kg s^-2 K^-1 

• Temperature is defined as  the degree and intensity of the heat  and its dimensional formula is written as;

            t = [M0L0T0K1]

• Distance is defined as the total movement of an object and  its dimensional formula is written as;

​           x =  [M0L1T0]

CALCULATION:

Given:

$P=\frac{\alpha }{\beta }{\log }_e\left(\frac{kt}{\beta x}\right)$      -----(1)

Here, we have α and β constants, x is distance; k is the Boltzmann constant and t is the temperature.

The dimensional formula of Boltzmann constant, k = [M¹L²T^-2K^-1]

The dimensional formula of P = [M⁰L⁰T⁰]

The dimensional formula of distance,x = [M⁰L¹T⁰]

α and β are constants.

and the dimensional formula of temperature, t = [M⁰L⁰T⁰K¹]

Now, let us put these values in equation (1) we have;

$P=\frac{\alpha }{\beta }{\log }_e\left(\frac{kt}{\beta x}\right)$

⇒ [M0L0T0] = $\frac{\alpha }{\beta }$ log_e $(\frac{[M^1{L}^2{T}^{-2}{K}^{-1}][M^0{L}^0{T}^0{K}^1]}{\beta [M^0{L}^1{T}^0]})$

⇒  [M0L0T0] = $\frac{\alpha }{\beta }$ loge $(\frac{[M^1{L}^2{T}^{-2}{K}^0]}{\beta [M^0{L}^1{T}^0]})$

⇒  [M0L0T0] = $\frac{\alpha }{\beta }$ loge $(\frac{[M^1{L}^1{T}^{-2}{K}^0]}{\beta })$

⇒ $\frac{\beta }{\alpha }$ [M0L0T0] = loge $(\frac{[M^1{L}^1{T}^{-2}{K}^0]}{\beta })$

Taking the exponential on both sides we get;

 loge $(\frac{[M^1{L}^1{T}^{-2}{K}^0]}{\beta })$ =  $\frac{\beta }{\alpha }$ [M0L0T0] 

⇒ $\frac{[M^1{L}^1{T}^{-2}{K}^0]}{\beta }$ = $e^{\frac{\beta }{\alpha }[M^0{L}^0{T}^0]}$

$e^{\frac{\beta }{\alpha }[M^0{L}^0{T}^0]}$ should be dimensionless therefore we have;

$\frac{\beta }{\alpha }$ =1

α = β      ------(2)

and $\frac{[M^1{L}^1{T}^{-2}{K}^0]}{\beta }$ =1

⇒ β  = [MLT^-2]

Now, from equation (2) we have;

→ α = [MLT-2]

Hence, option 3) is the correct answer.