Book Online Demo
Check Your IQ
Try Test

Courses

Dropper NEET CourseDropper JEE CourseClass - 12 NEET CourseClass - 12 JEE CourseClass - 11 NEET CourseClass - 11 JEE CourseClass - 10 Foundation NEET CourseClass - 10 Foundation JEE CourseClass - 10 CBSE CourseClass - 9 Foundation NEET CourseClass - 9 Foundation JEE CourseClass -9 CBSE CourseClass - 8 CBSE CourseClass - 7 CBSE CourseClass - 6 CBSE Course

Q.

In face centred cubic (fcc) crystal lattice, edge length is 400 pm. Find the diameter (in pm) of greatest sphere which can be fit into the interstitial void without distortion of lattice.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

answer is 117.2.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

For a face centered cubic crystal
a2=4r
r=24a=a22=400×24=100×1.414=141.4pm
For an octahedral void a = 2(r+R)
Diameter of sphere = 2R = a–2r=400–2×14
Diameter = 2R = 400 – 282.8 =117.2 pm

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
whats app icon
In face centred cubic (fcc) crystal lattice, edge length is 400 pm. Find the diameter (in pm) of greatest sphere which can be fit into the interstitial void without distortion of lattice.