Sodium metal crystallizes in a body centred cubic lattice with unit cell edge length of $4~\mathop A\limits^o$. The radius of sodium atom is __________ $\times ~10^{-1}$ $\mathop A\limits^o$ (Nearest integer)
Answer (integer)
17
Solution
In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by :
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$\sqrt{3}a = 4r$
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Given the unit cell edge length (a) of sodium metal as 4 Å :
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$a = 4 ~\mathop A\limits^o$
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We can now solve for the radius (r) of the sodium atom :
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$4r = \sqrt{3}a$
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$r = \frac{\sqrt{3}a}{4}$
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$r = \frac{\sqrt{3} \times 4 ~\mathop A\limits^o}{4}$
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$r = \sqrt{3} ~\mathop A\limits^o$
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Now, we can approximate the numerical value :
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$r \approx 1.732 ~\mathop A\limits^o$
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To express the radius as a multiple of 10⁻¹ $\mathop A\limits^o$ :
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$$r = 1.732 \times 10^{1} \times 10^{-1} ~\mathop A\limits^o = 17.32 \times 10^{-1} ~\mathop A\limits^o$$
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So, the radius of the sodium atom is 17.32 × 10⁻¹ $\mathop A\limits^o$.
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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