The force of interaction between two atoms is given by $F=\alpha \beta \exp \left(-\frac{x^{2}}{\alpha \mathrm{kt}}\right)$; where $x$ is the distance, k is the Boltzmann constant and T is temperature and $\alpha$ and $\beta$ are two constants. The dimension of $\beta$ is:

(B) $\mathrm{M}^{2} \mathrm{LT}^{-4}$

(A) $\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-4}$

(D) $\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-2}$

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Units & Measurement

The force of interaction between two atoms is given by

F = α β exp (- \frac{x ^{2}}{α kt})

; where

x

is the distance, k is the Boltzmann constant and T is temperature and

α

and

β

are two constants. The dimension of

β

is:

Answer: (B)

The argument of an exponential must be dimensionless, so

[x^{2}] = [α k T]

. As

k T

carries the dimensions of energy,

M L^{2} T^{- 2}

, we get

L^{2} = [α] M L^{2} T^{- 2}

, so

[α] = M^{- 1} T^{2}

. The force equals

α β

apart from the dimensionless exponential, so

[β] = \frac{[ F ]}{[ α ]} = \frac{M L T ^{- 2}}{M ^{- 1} T ^{2}} = M^{2} L T^{- 4}

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option