Electronic polarizability and relative permittivity of a dilute rare gas A rare gas has N atoms per m^3. If the electronic polarizability of a single atom is α_e (SI units), which relation correctly connects ε_r and α_e in the dilute-gas limit?

Difficulty: Medium

Correct Answer: ε0 * (ε_r − 1) = N * α_e

Explanation:


Introduction / Context:
In gases at low density, the macroscopic polarization is simply the sum of independent atomic dipoles induced by the field. Connecting single-atom polarizability to bulk permittivity reveals how ε_r approaches 1 and increases linearly with number density N.


Given Data / Assumptions:

  • Dilute, non-interacting gas of atoms (rare gas).
  • Electronic polarizability per atom: α_e (SI).
  • Linear response, low fields.


Concept / Approach:

The polarization P equals dipole moment per unit volume: P = N * α_e * E. Also, D = ε0 E + P = ε E = ε0 ε_r E. Therefore, ε0 (ε_r − 1) E = P = N α_e E, giving ε0 (ε_r − 1) = N α_e. This is the dilute-gas limit of the more general Clausius–Mossotti relation, which introduces local-field corrections at higher densities.


Step-by-Step Solution:

Write P = N α_e E.Use D = ε0 E + P = ε0 ε_r E.Equate: ε0 (ε_r − 1) E = N α_e E ⇒ ε0 (ε_r − 1) = N α_e.


Verification / Alternative check:

As N → 0, ε_r → 1, consistent with vacuum; the linear dependence on N confirms the low-density assumption.


Why Other Options Are Wrong:

  • Options a and b miss the vacuum term; d uses “−2”, which is unphysical in this context.


Common Pitfalls:

Mixing cgs and SI forms; forgetting that polarization measures excess over vacuum response.


Final Answer:

ε0 * (ε_r − 1) = N * α_e

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