Edge of unit cell of fcc Xe crystal is 620pm. what is the radius of Xe atom
Answer (1)
Student Expert
5th Jul, 2019
hello Pratiksha,
The given radius of FCC XENON Crystal is 620 pm.then the atomic radius of Xe is found by.
For FCC Crystal,
Edge=2*R*underroot of 2.
Edge=620pm........given
620=2*R*underroot of 2
620/2=R*1.414.............underoot 2=1.414
310=R*1.414
310/1.414=R
R=219.86 pm
hence atomic radius of xe is 219.86.
hope this may help you
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