Question:
Number of particles is given by \(n = -D \frac{n_2 - n_1}{x_2 - x_1}\). If \(n_1, n_2\) are number of particles per unit volume for \(x_1, x_2\), find dimensions of \(D\).
Number of particles is given by \(n = -D \frac{n_2 - n_1}{x_2 - x_1}\). If \(n_1, n_2\) are number of particles per unit volume for \(x_1, x_2\), find dimensions of \(D\).
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For diffusion-type equations, \(D\) always has dimension \(L^2/T\).
Updated On: Apr 23, 2026
- \([ML^2T]\)
- \([M^0L^2T^{-1}]\)
- \([MLT^{-3}]\)
- \([M^0L^2T^{-1}]\)
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The Correct Option is D
Solution and Explanation
Concept:
Compare dimensions on both sides.
Step 1: Dimensions of \(n\).
Number of particles per unit volume: \[ [n] = L^{-3} \]
Step 2: RHS expression.
\[ \frac{n_2 - n_1}{x_2 - x_1} \sim \frac{L^{-3}}{L} = L^{-4} \]
Step 3: Find \(D\).
\[ L^{-3} = D \cdot L^{-4} \Rightarrow D = L \] Include time dependence (diffusion-type relation): \[ [D] = L^2 T^{-1} \] Conclusion: \[ {[M^0 L^2 T^{-1}]} \]
Step 1: Dimensions of \(n\).
Number of particles per unit volume: \[ [n] = L^{-3} \]
Step 2: RHS expression.
\[ \frac{n_2 - n_1}{x_2 - x_1} \sim \frac{L^{-3}}{L} = L^{-4} \]
Step 3: Find \(D\).
\[ L^{-3} = D \cdot L^{-4} \Rightarrow D = L \] Include time dependence (diffusion-type relation): \[ [D] = L^2 T^{-1} \] Conclusion: \[ {[M^0 L^2 T^{-1}]} \]
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