Number of particles is given by n = -D(n2-n1)/(x2-x1) crossing a unit area perpendicular to X- axis in unit time, where n1 and n2 are number of particles per unit volume for the value of x meant to x2 and x1. Find dimensions of D called as diffusion constant

The correct answer is Option M⁰L²T⁻¹.

Let us find the dimensions of the diffusion constant $D$ step-by-step using dimensional analysis.

The given formula is:
$n=-D\frac{n_2-n_1}{x_2-x_1}$
Let us determine the dimensions of each term in the equation:

1. Dimension of $n$:
According to the problem statement, $n$ is the number of particles crossing a unit area perpendicular to the X-axis in unit time.
Therefore, the formula for $n$ is:
$n=\frac{\text{Number of particles}}{\text{Area}\times \text{Time}}$
Since "Number of particles" is a dimensionless quantity (dimension $1$), the dimension of $n$ is:
$\left[n\right]=\frac{1}{L^2\times T}=L^{-2}{T}^{-1}$

2. Dimension of $(n_2-n_1)$:
Here, $n_1$ and $n_2$ are the number of particles per unit volume.
Thus, the dimension of concentration $n_1$ (or $n_2$) is:
$\left[n_1\right]=\left[n_2\right]=\frac{\text{Number of particles}}{\text{Volume}}=\frac{1}{L^3}=L^{-3}$
Subtracting two quantities of the same dimension yields a quantity with the same dimension. Hence:
$[n_2-n_1]=L^{-3}$

3. Dimension of $(x_2-x_1)$:
The terms $x_1$ and $x_2$ represent positions along the X-axis, which have the dimension of length ($L$).
Therefore:
$[x_2-x_1]=L$

4. Finding the dimension of $D$:
We rearrange the original formula to solve for the dimension of $D$ (ignoring the negative sign as it is dimensionless):
$\left[D\right]=\frac{\left[n\right]\times [x_2-x_1]}{[n_2-n_1]}$
Substitute the respective dimensions into the equation:
$\left[D\right]=\frac{\left(L^{-2}{T}^{-1}\right)\times L}{L^{-3}}$
Simplify the numerator:
$\left[D\right]=\frac{L^{-1}{T}^{-1}}{L^{-3}}$
Now, simplify by moving $L^{-3}$ to the numerator:
$\left[D\right]=L^{-1-(-3)}{T}^{-1}=L^2{T}^{-1}$
Expressing this in standard mass ($M$), length ($L$), and time ($T$) dimensions:
$\left[D\right]=M^0{L}^2{T}^{-1}$