Answer
\[\underline{1.28\times {{10}^{11}}\text{ km}}\] and \[\underline{2.4\times {{10}^{19}}\text{ km}}\]
Work Step by Step
\[1\text{ pm}={{10}^{-12}}\ m\]
Diameter of a hydrogen atom in meters is \[212\times {{10}^{-12}}\ m\].
Now, \[1\text{ km}={{10}^{3}}m\].
The length in kilometers of a row of \[6.02\times {{10}^{23}}\] atoms is calculated as follows:
\[\left( 6.02\times {{10}^{23}}\text{ atoms} \right)\left( \frac{212\times {{10}^{-12}}\ \text{m}}{1\text{ atom}} \right)\left( \frac{1\text{ km}}{{{10}^{3}}\text{ m}} \right)=1.28\times {{10}^{11\text{ }}}\text{km}\]
Since \[1\text{ cm}={{10}^{-2}}\ m\], diameter of a ping pong ball in meters is \[4\times {{10}^{-2}}\ m\].
So, the length in kilometers of a row of \[6.02\times {{10}^{23}}\] ping pong balls is calculated as follows:
\[\left( 6.02\times {{10}^{23}}\text{ balls} \right)\left( \frac{4.0\times {{10}^{-2}}\text{ m}}{1\text{ ball}} \right)\left( \frac{1\text{ km}}{{{10}^{3}}\text{ m}} \right)=2.4\times {{10}^{19\text{ }}}\text{km}\]
The length of a row of hydrogen atoms is \[\underline{1.28\times {{10}^{11}}\text{ km}}\] and that of ping pong balls is \[\underline{2.4\times {{10}^{19}}\text{ km}}\].
