Given mass number of gold = 197, density of gold = 19.7g cm–3, Avogadro’s number = 6 × 1023. The radius of the gold atom is approximately.

m=(vol)(density)A×mp=43πr3×( den ) ∴197×1.67×10−27=43×3.14r3×19.7×103 ∴r3=1.67×34×3.14×10−30∴r3=4×10−30 ∴r≅1.5×10−10m

Given mass number of gold = 197, density of gold = 19.7g cm–3, Avogadro’s number = 6 × 1023. The radius of the gold atom is approximately.

m=(vol)(density)A×mp=43πr3×( den ) ∴197×1.67×10−27=43×3.14r3×19.7×103 ∴r3=1.67×34×3.14×10−30∴r3=4×10−30 ∴r≅1.5×10−10m

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Given mass number of gold = 197, density of gold = 19.7g cm^–3, Avogadro’s number = 6 × 10²³. The radius of the gold atom is approximately.

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1.5 × 10^–10 m

1.5 × 10^–8 m

1.5 × 10^–9 m

1.5 × 10^–12 m

answer is C.

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Detailed Solution

m=(vol)(density)
$\begin{array}{l} A \times m_{p} = \frac{4}{3} {πr}^{3} \times (den) ∴ 197 \times 1 .67 \times 10^{- 27} \\ = \frac{4}{3} \times 3 .14 r^{3} \times 19 .7 \times 10^{3} ∴ r^{3} = \frac{1 .67 \times 3}{4 \times 3 .14} \times 10^{- 30} \\ ∴ r^{3} = 4 \times 10^{- 30} ∴ r ≅ 1 .5 \times 10^{- 10} m \end{array}$