# Given that the atomic radius of silver is 145*"pm", how many silver atoms will span a length of 4.70*mm?

Sep 6, 2017

In effect they are asking how many lengths of $290 \cdot \text{pm}$ will fit into a length of $4.70 \cdot m m$....

#### Explanation:

Now we can do this dimensionally......

$\frac{4.70 \times {10}^{-} 3 \cdot m}{290 \times {10}^{-} 12 \cdot m} = 1.621 \times {10}^{7}$ individual silver atoms.....

Why did I double the given radius? Did I make an error?

We note that $1 \cdot m m \equiv 1 \times {10}^{-} 3 \cdot m$...

And that $1 \cdot \text{picometre} \equiv 1 \times {10}^{-} 12 \cdot m$

The rest is arithmetic. Of course, we have to do the ratio properly. I have never found it straightforward.