Question

In the relation $P=\frac{\alpha }{\beta }{e}^{\frac{-\alpha Z}{K\theta }}$ , $P$ is the pressure, $Z$ is the distance, $K$ is the Boltzmann constant and $\theta$ is the temperature. The dimensional formula of $\beta$ will be:

• A. $\left[M^{-1}{L}^2{T}^{-1}\right]$
• B. $\left[M^{-1}{L}^0{T}^{-1}\right]$
• C. $\left[M^0{L}^2{T}^0\right]$
• D. $\left[M^0{L}^2{T}^{-1}\right]$

Answer 1

The correct option is C $\left[M^0{L}^2{T}^0\right]$

Since the exponential function is a dimensionless quantity, therefore
$\left[\frac{\alpha Z}{K\theta }\right]=\left[M^0{L}^0{T}^0\right]$
$\left[\alpha \right]=\left[\frac{K\theta }{Z}\right]$
But $\left[P\right]=\left[\frac{\alpha }{\beta }\right]$
So, $\left[\beta \right]=\left[\frac{\alpha }{P}\right]=\left[\frac{K\theta }{ZP}\right]\dots \left(i\right)$
Dimension of Boltzmann constant $K=\left[m^1{L}^2{T}^{-2}{K}^{-1}\right]$
Dimension of temperature $\theta =\left[K^1\right]$
Dimension of distance $Z=\left[L^1\right]$
Dimension of pressure $P=\left[M^1{L}^{-1}{T}^{-2}\right]$
Now from equation $\left(i\right)$
$=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k^1\right]}{\left[M^1{L}^{-1}{T}^{-2}\right]\left[L^1\right]}=\left[M^0{L}^2{T}^0\right]$