2\sqrt{2}}\];   \[r=\frac{620}{2\sqrt{2}}=219.25\,PM\]

"> 2\sqrt{2}}\];   \[r=\frac{620}{2\sqrt{2}}=219.25\,PM\]

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The edge of unit cell of FCC \[Xe\] crystal is \[620\,pm\]. The radius of \[Xe\] atom is      [MP PET 2004]

Joint Entrance Exam - Main (JEE Main) Chemistry in Joint Entrance Exam - Main (JEE Main) 11 months ago

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\[r=\frac{a}{2\sqrt{2}}\];   \[r=\frac{620}{2\sqrt{2}}=219.25\,PM\]

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Posted on 30 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Chemistry in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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