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If the diameter of a carbon atom is \(0.15 \mathrm{~nm}\), calculate the number of carbon atoms which can be placed side by side in a straight line across the length of scale of length \(20 \mathrm{~cm}\) long.
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Diameter of carbon atom \(=0.15 \mathrm{~nm}\) \(=0.15 \times 10^{-9} \mathrm{~m}=1.5 \times 10^{-10} \mathrm{~m}\)
Length along which atoms are to be placed \(=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}=2 \times 10^{-1} \mathrm{~m}\)
\(\therefore\) No. of C-atoms which can be placed along the length \(=\frac{2 \times 10^{-1}}{1.5 \times 10^{-10}}=1.33 \times 10^9\)
Length along which atoms are to be placed \(=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}=2 \times 10^{-1} \mathrm{~m}\)
\(\therefore\) No. of C-atoms which can be placed along the length \(=\frac{2 \times 10^{-1}}{1.5 \times 10^{-10}}=1.33 \times 10^9\)
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