Question

The number of particles crossing unit area perpendicular to X-axis in unit time is given by $\frac{D(n_2-n_1)}{(x_2-x_1)}$ where $n_{1}and{n}_2$ are number of particle per unit volume respectively at and $x_2$ The dimension of the diffusion constant D are

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

Answer 1

Answer: (B)

$\left[L^2{T}^{-1}\right]$

$\begin{array}{c}\left[n\right]={\left[L\right]}^{-2}\times {\left[T\right]}^{-1}\\ \left[n_2-n_1\right]={\left[L\right]}^{-3}\\ \left[x_2-x_1\right]=\left[L\right]\\ \left[n\right]=\left[d\right]\times \left[n_2-n_1\right]/\left[x_2-x_1\right]\\ puttheabovevaluesin:-\\ {\left[L\right]}^{-2}\times {\left[T\right]}^{-1}=\left[d\right]\times {\left[L\right]}^{-3}/\left[L\right]\\ \left[d\right]={\left[L\right]}^{-2}\times {\left[T\right]}^{-1}\times \left[L\right]/{\left[L\right]}^{-3}\\ \left[d\right]={\left[L\right]}^{2+1+3}\times {\left[T\right]}^{-1}\\ \left[d\right]={\left[L\right]}^2\times {\left[T\right]}^{-1}=\left[L^2{T}^{-1}\right]\end{array}$

Hence,

option $\left(B\right)$ is correct answer.