Units & Dimensions

JEE Main level, single correct, medium

125 of 149

Question 125 of 149

The force of interaction between two atoms is given by F=αβexp⁡(−x2αkT);F=\alpha \beta \exp \left(-\frac{x^{2}}{\alpha k T}\right) ; where xx is the distance, k\mathrm{k} is the

Boltzmann constant and T is temperature and α\alpha and β\beta are two constants. The dimensions of β\beta is:

Show answer and solution

Answer: (B) M2LT−4\mathrm{M}^{2} \mathrm{LT}^{-4}

Force of interaction between two atoms,

F=αβe(−x2αkT)\mathrm{F}=\alpha \beta \mathrm{e}^{\left(\frac{-\mathrm{x}^{2}}{\alpha \mathrm{kT}}\right)}

Since exponential terms are dimensionless ∴[x2αkT]=M0 L0 T0\therefore\left[\frac{\mathrm{x}^{2}}{\alpha \mathrm{k} \mathrm{T}}\right]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}

⇒L2[α]ML2 T−2=M0 L0 T0\Rightarrow \frac{\mathrm{L}^{2}}{[\alpha] \mathrm{ML}^{2} \mathrm{~T}^{-2}}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}

⇒[α]=M−1 T2\Rightarrow[\alpha]=\mathrm{M}^{-1} \mathrm{~T}^{2}

[F]=[α][β][\mathrm{F}]=[\alpha][\beta]

MLT−2=M−1 T2[β]\mathrm{MLT}^{-2}=\mathrm{M}^{-1} \mathrm{~T}^{2}[\beta]

⇒[β]=M2LT−4\Rightarrow[\beta]=M^{2} L T^{-4}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Units & Dimensions (Dimensions), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.