Question

Number of particles is given by $n=D\frac{n_2-h_1}{x_2-x_1}$ crossing a unit area perpendicular to $X-\text{axis}$ in unit time, when $n_1$ and $n_2$ are number of particles per unit volume for the value of $x$ meant to $x_2$ and $x_1$. Find dimension of $D$ called as diffusion constant.

• A. $\left[M^{0}L{T}^2\right]$
• B. $\left[M^{0}L{T}^{-3}\right]$
• C. $\left[M^0{L}^2{T}^{-1}\right]$
• D. $\left[M^0{L}^2{T}^{-4}\right]$

Answer 1

The correct option is C $\left[M^0{L}^2{T}^{-1}\right]$

Given: $n=D\frac{n_2-h_1}{x_2-x_1}\dots \left(i\right)$
Where,
$\left[n\right]=$ Number of particles crossing a unit area in unit time $=\left[L^{-2}{T}^{-1}\right]$

$\left[n_2\right]=\left[n_1\right]=$ number of particles per unit volume $=\left[L^{-3}\right]$

$\left[x_2\right]=\left[x_1\right]=$ positions $=\left[L\right]$

Therefore, by putting the dimensions in equation $\left(i\right)$ we get,

$D=\frac{\left[n\right][x_2-x_1]}{[n_2-n_1]}$

$D=\frac{\left[L^{-2}{T}^{-1}\right]\times \left[L\right]}{L^{-3}}$

$D=\left[M^0{L}^2{T}^{-1}\right]$