JEE Challenger
More from Units and Measurements
PhyMedAdv 2016 • Paper 1

Dimensionally Correct Length Scale Expressions Involving Permittivity

Consider a characteristic length scale ll depending on the dielectric permittivity ε\varepsilon, Boltzmann's constant kBk_B, absolute temperature TT, particle number density nn, and charge qq of each particle. Which of the following expression(s) for ll is(are) dimensionally correct?

Options

A

l=(nq2εkBT)l = \sqrt{\left(\frac{n q^2}{\varepsilon k_B T}\right)}

B

l=(εkBTnq2)l = \sqrt{\left(\frac{\varepsilon k_B T}{n q^2}\right)}

Correct
C

l=(q2εn2/3kBT)l = \sqrt{\left(\frac{q^2}{\varepsilon n^{2/3} k_B T}\right)}

D

l=(q2εn1/3kBT)l = \sqrt{\left(\frac{q^2}{\varepsilon n^{1/3} k_B T}\right)}

Correct

Topics & Concepts

Units and MeasurementsDielectrics

Step-by-Step Solution

To determine which expression(s) for the length scale ll are dimensionally correct, we analyze the dimensions of each constituent physical quantity:

  1. Length (ll): [l]=L[l] = \text{L}

  2. Particle number density (nn): It is defined as the number of particles per unit volume. [n]=1Volume=L−3[n] = \frac{1}{\text{Volume}} = \text{L}^{-3}

  3. Thermal energy (kBTk_B T): Boltzmann's constant kBk_B multiplied by absolute temperature TT gives energy. [kBT]=Energy=M L2T−2[k_B T] = \text{Energy} = \text{M L}^2 \text{T}^{-2}

  4. Dielectric permittivity (ε\varepsilon): From Coulomb's Law, the electrostatic potential energy between two charges qq at a distance rr is given by U=q24πεrU = \frac{q^2}{4\pi \varepsilon r}. Rearranging for permittivity ε\varepsilon: ε=q24πUr  ⟹  [ε]=[q]2[kBT]⋅L\varepsilon = \frac{q^2}{4\pi U r} \implies [\varepsilon] = \frac{[q]^2}{[k_B T] \cdot \text{L}} Therefore, the dimensional combination [q2εkBT]\left[ \frac{q^2}{\varepsilon k_B T} \right] simplifies to: [q2εkBT]=L\left[ \frac{q^2}{\varepsilon k_B T} \right] = \text{L}

Now, let's evaluate the dimensions for each option:

Option A:

l=nq2εkBTl = \sqrt{\frac{n q^2}{\varepsilon k_B T}} Substituting the dimensions: [l]=[n]⋅[q2εkBT]=L−3⋅L=L−2=L−1[l] = \sqrt{[n] \cdot \left[\frac{q^2}{\varepsilon k_B T}\right]} = \sqrt{\text{L}^{-3} \cdot \text{L}} = \sqrt{\text{L}^{-2}} = \text{L}^{-1} Since [l]=L−1≠L[l] = \text{L}^{-1} \neq \text{L}, Option A is dimensionally incorrect.

Option B:

l=εkBTnq2l = \sqrt{\frac{\varepsilon k_B T}{n q^2}} Substituting the dimensions: [l]=1[n]⋅[εkBTq2]=1L−3⋅1L=L3⋅L−1=L2=L[l] = \sqrt{\frac{1}{[n]} \cdot \left[\frac{\varepsilon k_B T}{q^2}\right]} = \sqrt{\frac{1}{\text{L}^{-3}} \cdot \frac{1}{\text{L}}} = \sqrt{\text{L}^3 \cdot \text{L}^{-1}} = \sqrt{\text{L}^2} = \text{L} Since [l]=L[l] = \text{L}, Option B is dimensionally correct. (Note: This is the expression for the Debye screening length).

Option C:

l=q2εn2/3kBTl = \sqrt{\frac{q^2}{\varepsilon n^{2/3} k_B T}} The dimension of n2/3n^{2/3} is: [n2/3]=(L−3)2/3=L−2[n^{2/3}] = (\text{L}^{-3})^{2/3} = \text{L}^{-2} Substituting into the expression: [l]=[q2εkBT]⋅1[n2/3]=L⋅1L−2=L3=L3/2[l] = \sqrt{\left[\frac{q^2}{\varepsilon k_B T}\right] \cdot \frac{1}{[n^{2/3}]}} = \sqrt{\text{L} \cdot \frac{1}{\text{L}^{-2}}} = \sqrt{\text{L}^3} = \text{L}^{3/2} Since [l]=L3/2≠L[l] = \text{L}^{3/2} \neq \text{L}, Option C is dimensionally incorrect.

Option D:

l=q2εn1/3kBTl = \sqrt{\frac{q^2}{\varepsilon n^{1/3} k_B T}} The dimension of n1/3n^{1/3} is: [n1/3]=(L−3)1/3=L−1[n^{1/3}] = (\text{L}^{-3})^{1/3} = \text{L}^{-1} Substituting into the expression: [l]=[q2εkBT]⋅1[n1/3]=L⋅1L−1=L2=L[l] = \sqrt{\left[\frac{q^2}{\varepsilon k_B T}\right] \cdot \frac{1}{[n^{1/3}]}} = \sqrt{\text{L} \cdot \frac{1}{\text{L}^{-1}}} = \sqrt{\text{L}^2} = \text{L} Since [l]=L[l] = \text{L}, Option D is dimensionally correct.

Correct Options: B and D

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