A solid consisting of only X-atoms has a close-packed structure with X-X distance of 160 pm. Assuming it to be a closed packed structure of hard spheres with radius equal to half of the X-X bond length, the number of atoms in $1 \mathrm{cm}^{3}$ would be
(a)$6.023 \times 10^{27}$
(b)$3.45 \times 10^{23}$
(c)$6.02 \times 10^{21}$
(d)$3.8 \times 10^{21}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Close packing (fcc): a = √2 × 160 = 226 pm; volume per atom = a³/4 = 2.89×10⁻²⁴ cm³ ⇒ 3.46×10²³ atoms per cm³.