JEE PYQ: Solid State Question 8
Question 8 - 2020 (03 Sep Shift 1)
An element with molar mass $2.7 \times 10^{-2}$ kg mol$^{-1}$ forms a cubic unit cell with edge length 405 pm. If its density is $2.7 \times 10^3$ kg m$^{-3}$, the radius of the element is approximately ___ $\times 10^{-12}$ m (to the nearest integer).
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Answer: 143
Solution
$d = \dfrac{Z \times M}{N_A \times \text{Volume}}$. $2.7 = \dfrac{Z \times 27}{6.02 \times 10^{23} \times (4.05 \times 10^{-8})^3}$. $Z = 4 \Rightarrow fcc$ unit cell. For fcc, $4r = \sqrt{2},a$. $r = \dfrac{1.414 \times 405}{4} = 143.1675 \text{ pm} = 143.17 \text{ pm} \approx 143$.
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