Atomic structure

1. Jan 27, 2015

shreddinglicks

1. The problem statement, all variables and given/known data
A golf O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is formed from a 100-mm length of 1.5-mm-diameter wire. Calculate the number of gold atoms in the O-ring.

2. Relevant equations
Volume of cylinder = PI*h*r^2

3. The attempt at a solution
I got the volume of the cylinder:
PI*100*.75^2 = 176.7

I used the formula m = D*V:

density gold = 19.28 g/cm^3
amu gold = 196.97 grams

19.28*176.7 = 3406.776

I then multiplied by avogadro's number over amu.

3406.776 *(6.02x10^23 / 196.97) = 1.04x10^25 atoms

The answer should be 10.4x10^21 atoms. I'm thinking I messed up some unit conversion. I am unsure of where.

2. Jan 27, 2015

Quantum Defect

What are the units on your volume? What are the units on your density?

3. Jan 27, 2015

shreddinglicks

The density is g/cm^3

the volume is mm^2

I see the issue, how do I convert this? Is there a way to convert so I can cancel out a unit?

4. Jan 27, 2015

Quantum Defect

Volume should be cubic something. mm^2 is an area.

To convert a volume in mm^3 to cm^3 think about the following:

Conversion factor * X mm^3 = Y cm^3

The conversion factor needs to have units of cm^3 on the top and mm^3 on the bottom. What about [( 1 cm/ 10 mm)]^3 ? I.e. can you use the conversion between cm and mm to come up with a conversion factor between mm^3 and cm^3 ?

5. Jan 27, 2015

shreddinglicks

I see, so I have [176.7 mm^3] * [1cm/10mm]^3

basically

176.7 / 10^3 = .1767

6. Jan 27, 2015

Staff: Mentor

Numbers are OK, but in the final formula you again ignored units. That's the simplest way of making mistakes, as you have already seen. Don't do that.

7. Jan 27, 2015

shreddinglicks

.1767cm^3

8. Jan 27, 2015

Staff: Mentor

Now use this number to answer the problem.