Question

The number of particles crossing a unit area perpendicular to x-axis in a unit time is given by $n=-D\dfrac{{{n}_{2}}-{{n}_{1}}}{{{x}_{2}}-{{x}_{1}}}$, 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 diffusion constant. The dimensions of D are

Answer 1

Hint: In the above question, the expression for the number of particles crossing a unit area perpendicular to X-axis in unit time is given. First we need to express the quantities in the given expression in terms of their dimensions. Further accordingly substituting them in the above expression will enable us to determine the dimensions of D.

Formula used:
$\text{A=}\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{0}}} \right]$
$\text{V=}\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{3}}{{\text{T}}^{\text{0}}} \right]$
$\text{L=}\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{\text{0}}} \right]$
$t=\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{0}}\text{T} \right]$


Complete step-by-step solution:
In the above question it is given that ‘n’ is the number of particles crossing a unit area perpendicular to X-axis in unit time. The dimensions of area is $\text{A=}\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{0}}} \right]$ and that of time is $\text{t=}\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{0}}\text{T} \right]$. Hence the dimension n(d)of n is equal to,
$\begin{align}
  & n(d)=\dfrac{1}{A\times t} \\
 & \Rightarrow n(d)=\dfrac{1}{\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{0}}} \right]\times \left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{0}}\text{T} \right]} \\
 & \therefore n(d)=\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{-2}}}{{\text{T}}^{-1}} \right] \\
\end{align}$
The difference of a same physical quantity has the same dimension. In the question it is mentioned that ${{n}_{1}}$ and ${{n}_{2}}$ are the number of particles per unit volume. Hence the dimension ${{n}_{2}}-{{n}_{1}}(d)$ of ${{n}_{2}}-{{n}_{1}}$ is
$\begin{align}
  & {{n}_{2}}-{{n}_{1}}(d)=\dfrac{1}{V} \\
 & \Rightarrow {{n}_{2}}-{{n}_{1}}(d)=\dfrac{1}{\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{3}}{{\text{T}}^{\text{0}}} \right]} \\
 & \therefore {{n}_{2}}-{{n}_{1}}(d)=\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{-3}}{{\text{T}}^{\text{0}}} \right] \\
\end{align}$
Similarly the dimension ${{x}_{2}}-{{x}_{1}}(d)$ of ${{x}_{2}}-{{x}_{1}}$ is given by,
$\begin{align}
  & {{x}_{2}}-{{x}_{1}}(d)=L \\
 & \therefore {{x}_{2}}-{{x}_{1}}(d)=\text{L=}\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{\text{0}}} \right] \\
\end{align}$
From the relation given in the question, the dimension D(d) of diffusion constant is equal to,
$n(d)=-D(d)\dfrac{{{n}_{2}}-{{n}_{1}}(d)}{{{x}_{2}}-{{x}_{1}}(d)}$
$\Rightarrow \left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{-2}}}{{\text{T}}^{-1}} \right]=-D(d)\dfrac{\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{-3}}{{\text{T}}^{\text{0}}} \right]}{\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{\text{0}}} \right]}$
$\begin{align}
  & \Rightarrow \left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{-2}}}{{\text{T}}^{-1}} \right]=-D(d)\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{-4}}{{\text{T}}^{\text{0}}} \right] \\
 & \therefore D(d)=\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{-1}} \right] \\
\end{align}$
Therefore the dimension of D are $\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{-1}} \right]$


Note: Number of particles is basically a constant. Hence it does not have any dimension. It is also to be noted that we have ignored the negative sign in the above relation given. This is because a negative sign can be considered as -1. Since -1 is nothing but a constant, hence it can be implied that it is dimensionless. It is also to be noted that the powers of the fundamental dimensions of the above physical quantities are added or subtracted using the laws of exponent.