The force of interaction between two atoms is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The argument of the exponential function must be dimensionless.Therefore,$[\frac{x^2}{\alpha kt}] = [M^0L^0T^0]$.Since $[x^2] = L^2$ and $[kt] = [

Vedclass · Accepted Answer

$M^2LT^{-4}$

Vedclass · Answer

(B) The argument of the exponential function must be dimensionless.Therefore,$[\frac{x^2}{\alpha kt}] = [M^0L^0T^0]$.Since $[x^2] = L^2$ and $[kt] = [Energy] = ML^2T^{-2}$,we have:$[\alpha] = \frac{[x^2]}{[kt]} = \frac{L^2}{ML^2T^{-2}} = M^{-1}T^2$.The force $F$ is given by $F = \alpha \beta \exp(\dots)$. Since the exponential term is dimensionless,the dimensions of $F$ must be equal to the dimensions of $\alpha \beta$.$[F] = [\alpha][\beta]$$MLT^{-2} = (M^{-1}T^2) [\beta]$$[\beta] = \frac{MLT^{-2}}{M^{-1}T^2} = M^2LT^{-4}$.

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

Similar Questions

Is it possible for a physical quantity to have dimensions but no units?

What is the dimension of the ratio of Planck's constant $(h)$ to the moment of inertia $(I)$?

The dimensions of $(\mu_0 \varepsilon_0)^{-1/2}$ are

Dimensions of the coefficient of viscosity are

The de-Broglie wavelength associated with a particle of mass $m$ and energy $E$ is $h / \sqrt{2 m E}$. The dimensional formula for Planck's constant $h$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module