If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

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Solution

The diameter of 1 C atom is $0.15 n m = 0.15 \times 10^{- 7} c m$ .
In 20 cm length, $\frac{20}{0.15 \times 10^{- 7}} = 1.33 \times 10^{9}$ carbon atoms can be placed.

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If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

The diameter of 1 C atom is 0.15nm=0.15×10−7cm. In 20 cm length, 200.15×10−7=1.33×109 carbon atoms can be placed.

The diameter of 1 C atom is $0.15 n m = 0.15 \times 10^{- 7} c m$ .
In 20 cm length, $\frac{20}{0.15 \times 10^{- 7}} = 1.33 \times 10^{9}$ carbon atoms can be placed.