Question

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

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

Answer 1

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

Using principle of homogeneity
$\left[P\right]=\left[\frac{\alpha }{\beta }\right]\left[e^{-\frac{\alpha z}{k\theta }}\right]$
$\left[P\right]=\left[\frac{\alpha }{\beta }\right][\because [e^{-\frac{\alpha z}{k\theta }}]=1]$
$\left[M{L}^{-1}{T}^{-2}\right]=\left[\frac{\alpha }{\beta }\right]$
$\Rightarrow \left[\beta \right]=\frac{\left[\alpha \right]}{\left[M{L}^{-1}{T}^{-2}\right]}...\left(1\right)$
Since powers of exponentials should be dimensionless
$\left[\frac{\alpha z}{k\theta }\right]=1$
$\Rightarrow \left[\alpha \right]=\frac{\left[k\theta \right]}{\left[z\right]}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L\right]}[\because k\theta \text{has dimensions of energy}]$

$\left[\alpha \right]=\left[ML{T}^{-2}\right]...\left(2\right)$
Substituting $\left(2\right)$ in $\left(1\right)$
$\left[\beta \right]=\frac{ML{T}^{-2}}{\left[M{L}^{-1}{T}^{-2}\right]}=\left[M^0{L}^2{T}^0\right]$