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The work done by a gas molecule in an isolated system is given by, $W = α β^{2} e^{\frac{- x^{2}}{α k T}}$ , where x is the displacement k is the Boltzmann constant and T is the temperature, $α$ and $β$ are constants. Then the dimensions of $β$ will be :

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$[M^{0} L T^{0}]$

$[M^{2} L T^{2}]$

$[M L T^{- 2}]$

$[M L^{2} T^{- 2}]$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{x^{2}}{α k T} = dimensionless \\ \Rightarrow \frac{x^{2}}{k T} = α \\ Using, \\ v = \sqrt{\frac{3 k T}{m}} \\ \Rightarrow k T = \frac{v^{2} \cdot m}{3} \end{array}$

So,

$\Rightarrow α = \frac{x^{2}}{m v^{2}} = L^{2} M^{- 1} L^{- 2} T^{2} = M^{- 1} T^{2}$

$Now, given that: work = α \cdot β^{2} \cdot (dimensionless)$

$So, dimensions of work is same as dimensions of α \cdot β^{2}$

$\Rightarrow M^{1} L^{1} T^{- 2} \cdot L^{1} = M^{- 1} T^{2} β^{2}$

$\begin{array}{l} \Rightarrow β = M^{1} L^{1} T^{- 2} \end{array}$

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