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Question: A dimensionless quantity is constructed in terms of electronic charge e, permittivity of free space ...

A dimensionless quantity is constructed in terms of electronic charge e, permittivity of free space ϵ0\epsilon_0, Planck's constant h, and speed of light c. If the dimensionless quantity is written as eαϵ0βhγcδe^\alpha\epsilon_0^\beta h^\gamma c^\delta and n is a non-zero integer, then (α,β,γ,δ)(\alpha, \beta, \gamma, \delta) is given by :

A

(2n,−n,−n,−n)(2n, -n, -n, -n)

B

(n,−n,−2n,−n)(n, -n, -2n, -n)

C

(n,−n,−n,−2n)(n, -n, -n, -2n)

D

(2n,−n,−n,−2n)(2n, -n, -n, -2n)

Answer

(2n, -n, -n, -n)

Explanation

Solution

The dimensionless quantity is given by Q=eαϵ0βhγcδQ = e^\alpha\epsilon_0^\beta h^\gamma c^\delta.

The dimensions of the quantities are:

  • Electronic charge, ee: [IT]
  • Permittivity of free space, ϵ0\epsilon_0: [M−1L−3T4I2][M^{-1}L^{-3}T^4I^2] (from Coulomb's law F=14πϵ0q1q2r2F = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2})
  • Planck's constant, hh: [ML2T−1][ML^2T^{-1}] (from E=hfE=hf)
  • Speed of light, cc: [LT−1][LT^{-1}]

For Q to be dimensionless, its dimensions must be [M0L0T0I0][M^0L^0T^0I^0].

[Q]=[e]α[ϵ0]β[h]γ[c]δ[Q] = [e]^\alpha [\epsilon_0]^\beta [h]^\gamma [c]^\delta

[M0L0T0I0]=[IT]α[M−1L−3T4I2]β[ML2T−1]γ[LT−1]δ[M^0L^0T^0I^0] = [IT]^\alpha [M^{-1}L^{-3}T^4I^2]^\beta [ML^2T^{-1}]^\gamma [LT^{-1}]^\delta

[M0L0T0I0]=[M−β+γ][L−3β+2γ+δ][Tα+4β−γ−δ][Iα+2β][M^0L^0T^0I^0] = [M^{-\beta+\gamma}] [L^{-3\beta+2\gamma+\delta}] [T^{\alpha+4\beta-\gamma-\delta}] [I^{\alpha+2\beta}]

Equating the exponents of M, L, T, and I to zero:

  • M: −β+γ=0  ⟹  γ=β-\beta + \gamma = 0 \implies \gamma = \beta
  • L: −3β+2γ+δ=0-3\beta + 2\gamma + \delta = 0
  • T: α+4β−γ−δ=0\alpha + 4\beta - \gamma - \delta = 0
  • I: α+2β=0  ⟹  α=−2β\alpha + 2\beta = 0 \implies \alpha = -2\beta

Substitute γ=β\gamma = \beta into the L and T equations:

  • L: −3β+2β+δ=0  ⟹  −β+δ=0  ⟹  δ=β-3\beta + 2\beta + \delta = 0 \implies -\beta + \delta = 0 \implies \delta = \beta
  • T: α+4β−β−δ=0  ⟹  α+3β−δ=0\alpha + 4\beta - \beta - \delta = 0 \implies \alpha + 3\beta - \delta = 0

Substitute α=−2β\alpha = -2\beta and δ=β\delta = \beta into the last equation:

(−2β)+3β−β=0(-2\beta) + 3\beta - \beta = 0

β−β=0\beta - \beta = 0

0=00 = 0

This confirms consistency, and we have one free parameter. Let β=−n\beta = -n, where n is a non-zero integer as given in the question.

Then α=−2β=−2(−n)=2n\alpha = -2\beta = -2(-n) = 2n.

γ=β=−n\gamma = \beta = -n.

δ=β=−n\delta = \beta = -n.

Thus, (α,β,γ,δ)=(2n,−n,−n,−n)(\alpha, \beta, \gamma, \delta) = (2n, -n, -n, -n).