Problem 8
Question
The number of particles given by \(n=D \frac{n_{2}-n_{1}}{x_{2}-x_{1}}\) are 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 the values \(x_{1}\) and \(x_{2}\) of \(x\) respectively. Then the dimensional formula of diffusion constant \(D\) is (a) \(\left[\mathrm{M}^{0} \mathrm{LT}^{0}\right]\) (b) \(\left[\mathrm{M}^{0} \mathfrak{2}^{2} \mathrm{~T}^{-4}\right]\) (c) \(\left[\mathrm{M}^{0} \mathrm{LT}^{-3}\right]\) (d) \(\left[\mathrm{M}^{0}{\underline{\phantom{xx}}}^{2} \mathrm{~T}^{-1}\right]\)
Step-by-Step Solution
Verified Answer
The dimensional formula of the diffusion constant \( D \) is \([M^{0} L^{2} T^{-1}]\), which corresponds to option (d).
1Step 1: Identify known quantities
The given equation for the number of particles is \( n = D \frac{n_{2} - n_{1}}{x_{2} - x_{1}} \), where \( n \) is the number of particles crossing a unit area per unit time. We know that \( x_{2} - x_{1} \) represents a length dimension \( [L] \), and \( n_{2} - n_{1} \) represents the difference in concentration, which is particles per unit volume, \( [L^{-3}] \).
2Step 2: Define dimensions of known terms
The dimensions of the number of particles \( n \) per unit time and unit area is \( [L^{-2}][T^{-1}] \), as particles \( n \) are crossing an area \( [L^2] \) perpendicular to the \( x \)-axis per time \( T \). The concentration gradient \( \frac{n_{2} - n_{1}}{x_{2} - x_{1}} \) has dimensions \( [L^{-3}] \) \( [L]^{-1} = [L^{-4}] \).
3Step 3: Solve for the diffusion constant dimensions
By rearranging and identifying units in the original equation, we have \( D = n \cdot \frac{x_{2} - x_{1}}{n_{2} - n_{1}} \). Substituting dimensions: \[ D = [L^{-2}][T^{-1}] \times [L][L^{3}] = [L^{2}][T^{-1}] \]. Therefore, the dimension of \( D \) is \( [M^{0}L^{2}T^{-1}] \).
4Step 4: Match with provided options
The derived dimensional formula of \( D \) is \( [M^{0}L^{2}T^{-1}] \). Comparing this with given options, it matches option (d): \([M^{0} L^{2} T^{-1}]\).
Key Concepts
Dimensional AnalysisParticle FlowConcentration GradientPhysics Problem Solving
Dimensional Analysis
Dimensional analysis is a powerful tool in physics to understand the relationships between different physical quantities. It involves checking the consistency of equations with regard to the dimensions assigned to each physical quantity. Here, the term "dimension" refers to the measure of a physical quantity (like length, time, mass, etc.) expressed using a base set of dimensions, typically denoted as \([L], [M], [T]\).
Consider a problem where you have an equation linking the number of particles \(n\), a diffusion constant \(D\), and a concentration gradient. By knowing the dimensions of these elements, we can solve for the unknown dimensions of \(D\). For example, if \(n\) has dimensions of \([L^{-2}][T^{-1}]\), derived from particles crossing a unit area in unit time, and the concentration gradient has dimensions of \([L^{-4}]\), we can find that \(D\) must have the dimensions \([L^{2}][T^{-1}]\) to ensure the equation is dimensionally consistent.
Using dimensional analysis often helps in checking the viability of equations and understanding how changes in one variable might affect another.
Consider a problem where you have an equation linking the number of particles \(n\), a diffusion constant \(D\), and a concentration gradient. By knowing the dimensions of these elements, we can solve for the unknown dimensions of \(D\). For example, if \(n\) has dimensions of \([L^{-2}][T^{-1}]\), derived from particles crossing a unit area in unit time, and the concentration gradient has dimensions of \([L^{-4}]\), we can find that \(D\) must have the dimensions \([L^{2}][T^{-1}]\) to ensure the equation is dimensionally consistent.
Using dimensional analysis often helps in checking the viability of equations and understanding how changes in one variable might affect another.
Particle Flow
Particle flow refers to the movement of particles from one area to another, often due to differences in concentration. In the given problem, we're looking at how particles move across a unit area perpendicular to the x-axis over time. This flow is described by the relationship between the number of particles crossing this area, the diffusion constant, and the concentration gradient.
In practice, particle flow is greatly influenced by factors like temperature, pressure, and the properties of the medium in which the particles are moving. Understanding particle flow is crucial in many fields such as chemistry and physics, as it explains phenomena like diffusion, where particles move from regions of high concentration to low concentration.
In practice, particle flow is greatly influenced by factors like temperature, pressure, and the properties of the medium in which the particles are moving. Understanding particle flow is crucial in many fields such as chemistry and physics, as it explains phenomena like diffusion, where particles move from regions of high concentration to low concentration.
- Flow synchronization: Ensures efficient and steady particle movement.
- Mean velocity: Average speed of particle flow that is key for predicting diffusion rates.
Concentration Gradient
The concentration gradient in physics refers to the gradual change in the concentration of particles over a particular distance. In the given problem, it is represented as \(\frac{n_2-n_1}{x_2-x_1}\), showing the change in particle concentration between two points along the x-axis.
A steep concentration gradient indicates a rapid change in concentration over a small distance, often leading to faster particle movement. Conversely, a shallow gradient means changes occur slowly. Thus, the concentration gradient is a key driver of diffusion.
A steep concentration gradient indicates a rapid change in concentration over a small distance, often leading to faster particle movement. Conversely, a shallow gradient means changes occur slowly. Thus, the concentration gradient is a key driver of diffusion.
- Direction of movement: Particles generally move from high to low concentration.
- Diffusion rate: Higher gradients generally result in a quicker spread of particles.
Physics Problem Solving
Physics problem solving involves a step-by-step approach to tackling complex physics questions. It is about applying known concepts, formulating a plan, performing calculations, and interpreting results. Let's break down the approach with our given exercise:
Firstly, identify what the problem is asking. Here, we want to find the dimensional formula for the diffusion constant. Once that's clear, take stock of known quantities and their dimensions. For instance, understanding that the number of particles per unit time and area gives dimensions of \([L^{-2}][T^{-1}]\).
Next, relate these dimensions within the governing equation: \(n = D \frac{n_2-n_1}{x_2-x_1}\). You'll need to rearrange this equation to solve for your unknown: the diffusion constant \(D\). Finally, using dimensional analysis, calculate the required dimensions and match with the given options.
Firstly, identify what the problem is asking. Here, we want to find the dimensional formula for the diffusion constant. Once that's clear, take stock of known quantities and their dimensions. For instance, understanding that the number of particles per unit time and area gives dimensions of \([L^{-2}][T^{-1}]\).
Next, relate these dimensions within the governing equation: \(n = D \frac{n_2-n_1}{x_2-x_1}\). You'll need to rearrange this equation to solve for your unknown: the diffusion constant \(D\). Finally, using dimensional analysis, calculate the required dimensions and match with the given options.
- Understanding the problem: Break down complex questions.
- Simplifying solutions: Use known dimensions and logical reasoning.
Other exercises in this chapter
Problem 7
\(\left[\mathrm{ML}^{3} \mathrm{~T}^{-1} \mathbf{Q}^{-1}\right]\) is the dimensional formula of (a) resistance (b) resistivity (c) capacitance (d) conductivity
View solution Problem 7
Which of the following cannot be regarded as an essential characteristic of a unit of measurement? (a) Inaccessibility (b) Indenstructibility (c) Invariability
View solution Problem 8
Young modulus of steel is \(1.9 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\). When expressed in CGS units of dyne \(/ \mathrm{cm}^{2}\), it will be equal to \(
View solution Problem 9
The dimensional formula of \(\frac{1}{\varepsilon_{0}} \frac{e^{2}}{h c}\) is (a) \(\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \mathrm{~A}^{0}\right]\
View solution