In a typical combustion engine the workdone by a gas molecule is given by W = \alpha ^2\beta e^ - \beta x^2 \over 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 :

Units and Measurements PYQ — JEE Main 2021 · 26 Feb · Shift 1 · Q52 with Solution

View written solutionFree

Correct answer: A

Given expression

$W=\alpha^2\beta\, e^{-\beta x^2/(kT)}$

where $W$ is work done.
Use the exponential condition

The argument of an exponential must be dimensionless. So,

$\frac{\beta x^2}{kT}$

is dimensionless.
Find dimension of $\beta$

We know:
- $[x]=L$
- $[kT]=$ energy $=[ML^2T^{-2}]$
Hence,

$[\beta][x^2]=[kT]$

$[\beta]L^2=ML^2T^{-2}$

Therefore,

$[\beta]=[MT^{-2}]$
Now use dimension of work

Since exponential is dimensionless,

$[W]=[\alpha^2\beta]$

Work has dimension:

$[W]=[ML^2T^{-2}]$

So,

$[\alpha^2][\beta]=ML^2T^{-2}$

Substituting $[\beta]=MT^{-2}$ ,

$[\alpha^2](MT^{-2})=ML^2T^{-2}$

$[\alpha^2]=L^2$

$[\alpha]=L$
Final dimension of $\alpha$

$[\alpha]=[M^0LT^0]$
Check options
- A: $[M^0LT^0]$ ✅
- B: $[MLT^{-1}]$ ❌
- C: $[MLT^{-2}]$ ❌
- D: $[M^2LT^{-2}]$ ❌

Therefore, the correct option is A.