In a typical combustion engine the work done by a gas molecule is given by $\mathrm{W}=\alpha^{2} \beta \mathrm{e}^{-\frac{\beta \mathrm{x}^{2}}{\mathrm{kT}}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be :

(B) $\left[\mathrm{M}^{0} \mathrm{~L} \mathrm{~T}^{0}\right]$

(A) $\left[\mathrm{M} \mathrm{LT}^{-1}\right]$

(C) $\left[\mathrm{M}^{2} \mathrm{~L} \mathrm{~T}^{-2}\right]$

(D) $\left[\mathrm{MLT}^{-2}\right]$

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Units & Measurement

In a typical combustion engine the work done by a gas molecule is given by

W = α^{2} β e^{- \frac{β x ^{2}}{kT}}

, where

x

is the displacement,

k

is the Boltzmann constant and

T

is the temperature. If

α

and

β

are constants, dimensions of

α

will be :

Answer: (B)

The exponent must be a pure number, so

β x^{2}

carries the dimensions of

k T

, that is of energy:

[β] = \frac{[ M L ^{2} T ^{- 2} ]}{[ L ^{2} ]} = [M T^{- 2}]

The exponential itself is dimensionless, and

W

is work, so

[α^{2} β] = [M L^{2} T^{- 2}]

and

[α^{2}] = \frac{[ M L ^{2} T ^{- 2} ]}{[ M T ^{- 2} ]} = [L^{2}]

α

has the dimensions of length,

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

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option