The Cu metal crystallises into fcc lattice with a unit cell edge length of 361 pm. The radius of Cu atom is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 127 pm
In a face-centred cubic (fcc) unit cell, the atoms touch each other along the face diagonal: the diagonal passes through one corner atom, the atom at the face centre and the opposite corner atom, so its length is r + 2r + r = 4r. By Pythagoras, the face diagonal of a cube of edge a is \( \sqrt{2}\,a \):
\( \displaystyle 4r=\sqrt{2}\,a\quad\Rightarrow\quad r=\frac{a}{2\sqrt{2}} \)
With a = 361 pm:
\( \displaystyle r=\frac{361\ \mathrm{pm}}{2\times1.414}=\frac{361\ \mathrm{pm}}{2.828}=127.6\ \mathrm{pm} \)
So the radius of the copper atom is about 127 pm.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Packing Efficiency in hcp and ccp Structures
