# Thread: [SOLVED] conversions (word problem)

1. ## [SOLVED] conversions (word problem)

The diameter of a neutral helium atom is about 1.0 102 pm. Suppose that we could line up helium atoms side by side in contact with one another. Approximately how many helium atoms would it take to make the distance from end to end 1 cm?

I am having problems setting the problem up. I know it wwould be a conversion problem.. and i know how to solve it from there, i just cant set it up.

2. Originally Posted by Steph07
The diameter of a neutral helium atom is about 1.0 102 pm. Suppose that we could line up helium atoms side by side in contact with one another. Approximately how many helium atoms would it take to make the distance from end to end 1 cm?

I am having problems setting the problem up. I know it wwould be a conversion problem.. and i know how to solve it from there, i just cant set it up.
Well one centimeter is one hundredth of a meter.

One picometer is $1 \times 10^{-12} m$

So $\frac{1cm}{1pm} = \frac{1 \times 10^{-2}}{ 1 \times 10^{-12}} = 10^{10}$

3. so it would take 10^10 atoms, would i put that back into scientific notation?

4. Originally Posted by Steph07
so it would take 10^10 atoms, would i put that back into scientific notation?
What do you mean by scientific notation?

There are no 'prefixes' for $\times 10^{10}$ I'm afraid.

It goes from Giga, which is $\times 10^9$ to Tera, which is $\times 10^{12}$

### how many helium atoms in 1 cm

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