If the diameter of a carbon atom is 0.15 nm,

Solution:

$1 \mathrm{~m}=100 \mathrm{~cm}$

$1 \mathrm{~cm}=10^{-2} \mathrm{~m}$

Length of the scale $=20 \mathrm{~cm}$

$=20 \times 10^{-2} \mathrm{~m}$

Diameter of a carbon atom $=0.15 \mathrm{~nm}$

$=0.15 \times 10^{-9} \mathrm{~m}$

One carbon atom occupies $0.15 \times 10^{-9} \mathrm{~m}$.

$\therefore$ Number of carbon atoms that can be placed in a straight line

$=\frac{20 \times 10^{-2} \mathrm{~m}}{0.15 \times 10^{-9} \mathrm{~m}}$

$=133.33 \times 10^{7}$

$=1.33 \times 10^{9}$