Question

Number of particles is given by $n=-D\dfrac{{{n}_{2}}-{{n}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ crossing a unit area perpendicular to X-axis in unit time, where ${{n}_{1}}$ and ${{n}_{2}}$ are the number of particles per unit volume for value of x meant to ${{x}_{2}}$ and ${{x}_{1}}$ . Find the dimensions of D called as diffusion constant:
a) $\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{\text{2}}} \right]$
b) $\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{2}}{{\text{T}}^{-4}} \right]$
c)$\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{-3}} \right]$
d)$\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{2}}{{\text{T}}^{-1}} \right]$

Answer 1

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 answer:
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,
$\begin{aligned} & n(d)=\dfrac{1}{A\times t} \\\ & \Rightarrow n(d)=\dfrac{1}{\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{0}}} \right]\times \left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{0}}\text{T} \right]} \\\ & \therefore n(d)=\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{-2}}}{{\text{T}}^{-1}} \right] \\\ \end{aligned}$
The difference of the 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
$\begin{aligned} & {{n}_{2}}-{{n}_{1}}(d)=\dfrac{1}{V} \\\ & \Rightarrow {{n}_{2}}-{{n}_{1}}(d)=\dfrac{1}{\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{3}}{{\text{T}}^{\text{0}}} \right]} \\\ & \therefore {{n}_{2}}-{{n}_{1}}(d)=\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{-3}}{{\text{T}}^{\text{0}}} \right] \\\ \end{aligned}$
Similarly the dimension ${{x}_{2}}-{{x}_{1}}(d)$ of ${{x}_{2}}-{{x}_{1}}$ is given by,
$\begin{aligned} & {{x}_{2}}-{{x}_{1}}(d)=L \\\ & \therefore {{x}_{2}}-{{x}_{1}}(d)=\text{L=}\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{\text{0}}} \right] \\\ \end{aligned}$
From the relation given in the question, the dimension D(d) of diffusion constant is equal to,
$\begin{aligned} & n(d)=-D(d)\dfrac{{{n}_{2}}-{{n}_{1}}(d)}{{{x}_{2}}-{{x}_{1}}(d)} \\\ & \Rightarrow \left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{-2}}}{{\text{T}}^{-1}} \right]=-D(d)\dfrac{\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{-3}}{{\text{T}}^{\text{0}}} \right]}{\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{\text{0}}} \right]} \\\ & \Rightarrow \left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{-2}}}{{\text{T}}^{-1}} \right]=-D(d)\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{-4}}{{\text{T}}^{\text{0}}} \right] \\\ & \therefore D(d)=\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{-1}} \right] \\\ \end{aligned}$

So, the correct answer is “Option d”.

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 negative signs 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.