JEE Challenger
More from Units and Measurements
PhyMedAdv 2024 • Paper 1

Dimensionless Quantity Combination of Fundamental Constants

A dimensionless quantity is constructed in terms of electronic charge ee, permittivity of free space ε0\varepsilon_0, Planck's constant hh, and speed of light cc. If the dimensionless quantity is written as eαε0βhγcδe^\alpha \varepsilon_0^\beta h^\gamma c^\delta and nn is a non-zero integer, then (α,β,γ,δ)(\alpha, \beta, \gamma, \delta) is given by

Options

A

(2n,−n,−n,−n)(2n, -n, -n, -n)

Correct
B

(n,−n,−2n,−n)(n, -n, -2n, -n)

C

(n,−n,−n,−2n)(n, -n, -n, -2n)

D

(2n,−n,−2n,−2n)(2n, -n, -2n, -2n)

Topics & Concepts

Units and MeasurementsCoulomb's Law

Step-by-Step Solution

To find the exponents (α,β,γ,δ)(\alpha, \beta, \gamma, \delta) such that the combination eαε0βhγcδe^\alpha \varepsilon_0^\beta h^\gamma c^\delta is dimensionless, we first write down the fundamental dimensions of each physical quantity in terms of Mass (M\text{M}), Length (L\text{L}), Time (T\text{T}), and Electric Current (A\text{A}):

  1. Electronic charge ee: [e]=AT[e] = \text{A} \text{T}

  2. Permittivity of free space ε0\varepsilon_0: [ε0]=M−1L−3T4A2[\varepsilon_0] = \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2

  3. Planck's constant hh: [h]=ML2T−1[h] = \text{M} \text{L}^2 \text{T}^{-1}

  4. Speed of light cc: [c]=LT−1[c] = \text{L} \text{T}^{-1}

The given quantity is Q=eαε0βhγcδQ = e^\alpha \varepsilon_0^\beta h^\gamma c^\delta. Substituting the dimensional formulas:

[Q]=(AT)α(M−1L−3T4A2)β(ML2T−1)γ(LT−1)δ[Q] = (\text{A} \text{T})^\alpha (\text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2)^\beta (\text{M} \text{L}^2 \text{T}^{-1})^\gamma (\text{L} \text{T}^{-1})^\delta

Grouping terms by base dimensions:

[Q]=M−β+γ⋅L−3β+2γ+δ⋅Tα+4β−γ−δ⋅Aα+2β[Q] = \text{M}^{-\beta + \gamma} \cdot \text{L}^{-3\beta + 2\gamma + \delta} \cdot \text{T}^{\alpha + 4\beta - \gamma - \delta} \cdot \text{A}^{\alpha + 2\beta}

For QQ to be dimensionless, the exponent of each dimension must be zero:

  1. For M\text{M}: −β+γ=0  ⟹  γ=β-\beta + \gamma = 0 \implies \gamma = \beta

  2. For A\text{A}: α+2β=0  ⟹  α=−2β\alpha + 2\beta = 0 \implies \alpha = -2\beta

  3. For L\text{L}: −3β+2γ+δ=0-3\beta + 2\gamma + \delta = 0 Substituting γ=β\gamma = \beta: −3β+2β+δ=0  ⟹  δ=β-3\beta + 2\beta + \delta = 0 \implies \delta = \beta

  4. For T\text{T}: α+4β−γ−δ=0\alpha + 4\beta - \gamma - \delta = 0 Substituting α=−2β\alpha = -2\beta, γ=β\gamma = \beta, and δ=β\delta = \beta: −2β+4β−β−β=0(Self-consistent)-2\beta + 4\beta - \beta - \beta = 0 \quad \text{(Self-consistent)}

Let β=−n\beta = -n, where nn is a non-zero integer. Then: α=−2(−n)=2n\alpha = -2(-n) = 2n β=−n\beta = -n γ=−n\gamma = -n δ=−n\delta = -n

Thus, the required tuple is: (α,β,γ,δ)=(2n,−n,−n,−n)(\alpha, \beta, \gamma, \delta) = (2n, -n, -n, -n)

This corresponds to Option (A).