The interionic distance for caesium chloride crystal will be
Chemistry · JEE Advanced · NTA Exams — Solid State
The interionic distance for caesium chloride crystal will be
- a
- \(\frac{a}{2}\)
- \(\frac{\sqrt{3} \mathrm{a}}{2}\)
- \(\frac{2 \mathrm{a}}{\sqrt{3}}\)
Answer
(C) 3 a 2
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