Question

the number of particles crossing a unit area perpendicular to x axis in unit time is given by n = -D(n2-n1/x2-x1) where n1 and n2 are number of particles per unit volume for the values of x meant to be x1 and x2 find the dimension of the diffusion constant D

Answer 1

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Answer 2

Answer:  

The dimension of diffusion constant D is $\bold{\left[\mathbf{L}^{2}\right]\left[\mathbf{T}^{-1}\right]}$.

Solution:

As the name suggests, the diffusion is defined as the number of particles crossing a given area per unit time. The dimensional constant has a unit of area per unit time i.e., $m^{2} s^{-1}$.

So, the dimension of it will be $\left[\mathrm{L}^{2}\right]\left[\mathrm{T}^{-1}\right] \text { and }[\mathrm{L}]$ is for metre and [T] is for second. This is because as  

$n=-D \times\left(\frac{n 2-n 1}{x 2-x 1}\right)$

So, the diffusion constant will be

$D=-n \times\left(\frac{x 2-x 1}{n 2-n 1}\right)$

The unit of the term (x2-x1) is metre and the term n is $m^{-2} s^{-1}.$ The unit of term $(n2-n1)\ is\ m^-3$. So inserting the units in above equation, we will get

$D=-m^{-2} s^{-1} \times\left(\frac{m}{m^{-3}}\right)=m^{-2+1+3} s^{-1}$

Thus,

$\bold{D=m^{2} s^{-1}=\left[L^{2}\right]\left[T^{-1}\right]}$