The number of particles crossing a unit area perpendicular to the $x$-axis in unit time is given by $n = -D\fr

Question

The number of particles crossing a unit area perpendicular to the $x$-axis in unit time is given by $n = -D\frac{n_2 - n_1}{x_2 - x_1}$, where $n_1$ and $n_2$ are the number of particles per unit volume for the value of $x$ equal to $x_1$ and $x_2$ respectively. The dimensions of $D$, known as the diffusion constant, will be:

Accepted Answer

1. **Find the dimensions of $n$:** $n$ is the number of particles crossing a unit area per unit time. 
Dimensions of $n = \frac{1}{	ext{Area} 	imes 	ext{Time}} = \frac{1}{[L^2][T]} = [L^{-2} T^{-1}]$.

2. **Find the dimensions of $(n_2 - n_1)$:** $n_1$ and $n_2$ are the number of particles per unit volume. 
Dimensions of $(n_2 - n_1) = \frac{1}{	ext{Volume}} = \frac{1}{[L^3]} = [L^{-3}]$.

3. **Find the dimensions of $(x_2 - x_1)$:** $x_1$ and $x_2$ are positions (distance). 
Dimensions of $(x_2 - x_1) = [L]$.

4. **Calculate dimensions of $D$:** Rearranging the given formula, we get $D = -n \frac{x_2 - x_1}{n_2 - n_1}$. 
Ignoring the negative sign for dimensions:
$[D] = \frac{[n] 	imes [x_2 - x_1]}{[n_2 - n_1]}$
$[D] = \frac{[L^{-2} T^{-1}] 	imes [L]}{[L^{-3}]} = \frac{[L^{-1} T^{-1}]}{[L^{-3}]} = [L^2 T^{-1}]$.

Therefore, the dimensional formula for $D$ is $[M^0 L^2 T^{-1}]$.

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