Question

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

[JEE(Main)-2021]

• A. [ML^2 T^-2]
• B. [MLT^-2]
• C. [M^2 LT^2]
• D. [M^0 L T^0]

Answer 1

Answer: (B)

[MLT^-2]

Answer 2

Given the work done,

$W = \alpha \beta^2 e^{-\frac{x^2}{\alpha kT}}$,

we note that:

• $W$ has dimensions of energy, i.e., $[W] = ML^2T^{-2}$.
• The exponential term must be dimensionless, so the exponent $\frac{x^2}{\alpha kT}$ is dimensionless.

Since $x$ has dimensions of length $[x]=L$ and $kT$ has dimensions of energy $ML^2T^{-2}$, we have:

$\frac{x^2}{\alpha kT}$ is dimensionless $\Rightarrow [\alpha] = \frac{L^2}{ML^2T^{-2}} = M^{-1}T^2$.

Now, from the work expression:

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

we substitute the dimensions:

$ML^2T^{-2} = (M^{-1}T^2)[\beta]^2$.

Thus,

$[\beta]^2 = \frac{ML^2T^{-2}}{M^{-1}T^2} = M^{2}L^2T^{-4}$.

Taking the square root, we get:

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