Question

The number of particles crossing per unit area perpendicular to $Z$ axis per unit time is given by $N=-D \frac{\left(N_{2}-N_{1}\right)}{\left(Z_{2}-Z_{1}\right)}$, where $N_{2}$ and $N_{1}$ are the number of particles per unit volume at $Z_{2}$ and $Z_{1}$ respectively. What is the dimensional formula for $D$ ?

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

Answer 1

Answer: (C)

$N=-D \frac{\left(N_{2}-N_{1}\right)}{\left(Z_{2}-Z_{1}\right)}$
Dimensionally,
$D=\frac{N\left(Z_{2}-Z_{1}\right)}{\left(N_{2}-N_{1}\right)}$
Given,
$N_{2},\, N_{1} \rightarrow$ Number of particles per unit volume.
$N_{2},\, N_{1} \rightarrow \frac{N}{V} \Rightarrow \left[L^{-3}\right]$
$Z_{2}-Z_{1} \rightarrow[L]$
$N \rightarrow \frac{\text { Number of particles }}{\text { Area } \cdot(T)}$
$N \rightarrow\left[L^{-2} T^{-1}\right]$
So, $D=\frac{L^{-2} T^{-1} \times L}{L^{-3}}$
$\Rightarrow \left[ L ^{2} T ^{-1}\right]$