Question

The number of particles crossing a unit area perpendicular to the x-axis in a unit area perpendicular to the $x-axis$ in a 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 at $x=x_1$ and $x=x_2$, respectively, and D is the diffusion constant. The dimensions of D are

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

Answer 1

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

Given,

$n=-D\left(\frac{n_2-n_1}{x_2-x_1}\right)$

Dimensional formula of $n=\left[L^{-2}{T}^{-1}\right]$

Dimensional formula of $n_1$ and $n_2=\frac{1}{\left[L^3\right]}=\left[L^{-3}\right]$

Dimensional formula of $x_1$ and $x_2=\left[L^{-1}\right]$

Now, substitute these values in the given equation

$D=n\frac{x_2-x_1}{n_2-n_1}$

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

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

Hence the dimensional formula of $D=\left[M^0{L}^2{T}^{-1}\right]$