Difficulty: Easy
Correct Answer: x ≪ R
Explanation:
Introduction / Context:
Electronic polarization in inert gases (He, Ne, Ar, etc.) arises from a very small distortion of the electron cloud relative to the nucleus when an external electric field is applied. This produces an induced dipole moment but only minute displacements in atomic length scales.
Given Data / Assumptions:
Concept / Approach:
The induced dipole moment p is proportional to the electronic polarizability αe and the local field: p = αe E. For a simple classical picture, p ≈ q x, where q is of the order of the total bound charge involved in the slight displacement. Because αe is small in rare gases, the resulting x is a tiny fraction of R, i.e., x ≪ R.
Step-by-Step Solution:
Recognize rare gases lack permanent dipoles; only electronic polarization occurs.Electronic polarizability values (in SI) are small, leading to very small induced dipole moments for modest E.Relate p to a hypothetical charge separation q over distance x: small p ⇒ very small x compared to atomic radius.Conclude x ≪ R.
Verification / Alternative check:
Order-of-magnitude estimates using αe for argon yield sub-picometer displacements under typical laboratory fields, far smaller than R ~ 0.1 nm.
Why Other Options Are Wrong:
Displacements comparable to R (0.5R, 0.75R, R) would imply enormous polarization and likely ionization or breakdown—physically unrealistic for small applied fields.
Common Pitfalls:
Overestimating atomic distortions; confusing macroscopic polarization with atomic-scale displacements.
Final Answer:
x ≪ R
Discussion & Comments