wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is 356 pm, then the radius (in pm) of carbon atom is:

A
77.07
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
154.14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
251.70
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
89.20
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 77.07
In a diamond structure, there are atoms at each corner and alternate tetrahedral voids. Consider the body diagonal of unit cell (cube with side a). It corresponds to 8R, and length of body diagonal is 3×a.
Thus R=3×a8=356×1.7328=77.07 pm

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Density
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon