1.45 Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
1.45 Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
1.45 For fcc unit cell, a=2√2?.
Here, a is the edge length and r is the atomic radius (0.144 nm).
a = 2√2 *0.144 = 0.407 nm
Hence, the length of a side of a cell is 0.407 nm.
Upvote 
Similar Questions for you
ΔG° = –RT * 2.303 log K
–nFE° = +RT * 2.303 log K
2 * 96500 * 0.295 = 8.314 * 298 * 2.303 log10 K
10 = log10 K = 1010
It has chiral centre and differently di substituted double bonded carbon atoms.
For FCC lattice
Packing efficiency = 
CsCl has BCC structure in which Cl– is present at corners of cube and Cs+ at body centre
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
Learn more about...
Chemistry Ncert Solutions Class 12th 2023
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering