Question

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.

Answer 1

1 m = 100 cm
1 cm = ${10}^{–2}$ m
Length of the scale = 20 cm
= 20 × ${10}^{–2}$ m
Diameter of a carbon atom = 0.15 nm
= 0.15 × ${10}^{–9}$ m
One carbon atom occupies 0.15 × ${10}^{–9}$ m.
$\therefore$ Number of carbon atoms that can be placed in a straight line
$=\frac{20\times {10}^{-2}m}{0.15\times {10}^{-9}m}=133.33\times {10}^7=1.33\times {10}^9$