Units & Dimensions

JEE Main level, single correct, medium

111 of 149

Question 111 of 149

The work done by a gas molecule in an isolated system is given by, W=αβ2e−x2αkT,W=\alpha {\beta }^{2}{e}^{-\dfrac{{x}^{2}}{\alpha kT}}, where xx is the displacement, kk is the Boltzmann constant and TT is the temperature. α\alpha and β\beta are constants. Then the dimensions of β\beta will be:

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Answer: (C) [MLT−2]\left[{MLT}^{-2}\right]

x2αkT→\dfrac{{x}^{2}}{\alpha kT}\rightarrow dimensionless
⇒[α]=[x2][kT]=L2ML2T−2=M−1T2\Rightarrow \left[\alpha \right]=\dfrac{\left[{x}^{2}\right]}{\left[kT\right]}=\dfrac{{L}^{2}}{{ML}^{2}{T}^{-2}}={M}^{-1}{T}^{2}
Now [W]=[α][β]2\left[W\right]=\left[\alpha \right]{\left[\beta \right]}^{2}
[β]=ML2T−2M−1T2=M1L1T−2\left[\beta \right]=\sqrt{\dfrac{{ML}^{2}{T}^{-2}}{{M}^{-1}{T}^{2}}}={M}^{1}{L}^{1}{T}^{-2}

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