Question

The number of particles crossing a unit cross-sectional area perpendicular to $x-$axis in unit time is given by $n=-D\left(\frac{n_2-n_1}{x_2-x_1}\right)$, where $n_1$ and $n_2$ are the number of particles per unit volume crossing a cross-sectional area at distances $x_2$ and $x_1$. Find the dimension of diffusion constant $D$.

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

Answer 1

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

Dimensional formula of $n$,
$\left[n\right]=\left[L^{-2}{T}^{-1}\right]$
Dimensional formula of $n_1$ and $n_2$,
$\left[n_1\right]=\left[n_2\right]=\left[L^{-3}\right]$
Dimensional formula of $x_1$ and $x_2$,
$\left[x_1\right]=\left[x_2\right]=\left[L\right]$
Given,
$n=-D\left(\frac{n_2-n_1}{x_2-x_1}\right)$
$\Rightarrow D=\frac{-n(x_2-x_1)}{(n_2-n_1)}$
$\Rightarrow \left[D\right]=\left[\frac{-n(x_2-x_1)}{(n_2-n_1)}\right]$
$\Rightarrow \left[D\right]=\left[\frac{L^{-2}{T}^{-1}\times L}{L^{-3}}\right]=\left[M^0{L}^2{T}^{-1}\right]$