Gold Atom Count in O-Ring: 10.4x10^21

In summary, the number of gold atoms in the golf O-ring can be calculated by first converting the volume from mm^3 to cm^3 using the conversion factor [1 cm/10 mm]^3 and then using the formula m = D*V to calculate the mass of the O-ring. Finally, multiplying the mass by Avogadro's number over the atomic mass of gold gives the number of gold atoms, which is 1.04x10^25 atoms. However, the correct answer should be 10.4x10^21 atoms, indicating a possible error in unit conversion.
  • #1
shreddinglicks
212
6

Homework Statement


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.

Homework Equations


Volume of cylinder = PI*h*r^2

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.
 
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  • #2
shreddinglicks said:

Homework Statement


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.

Homework Equations


Volume of cylinder = PI*h*r^2

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.

What are the units on your volume? What are the units on your density?
 
  • #3
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
shreddinglicks said:
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?

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

basically

176.7 / 10^3 = .1767
 
  • #6
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
.1767cm^3
 
  • #8
Now use this number to answer the problem.
 

1. What is the significance of the "Gold Atom Count" in an O-Ring?

The "Gold Atom Count" refers to the number of gold atoms present in the O-Ring. Gold is often used in scientific experiments as a marker or tracer, so knowing the number of gold atoms can help researchers track the movement or behavior of the O-Ring.

2. How is the "Gold Atom Count" measured in an O-Ring?

The "Gold Atom Count" can be measured using various methods such as X-ray fluorescence or atomic absorption spectroscopy. These techniques involve bombarding the O-Ring with x-rays or other forms of electromagnetic radiation and measuring the resulting emissions to determine the number of gold atoms present.

3. Is 10.4x10^21 a high or low "Gold Atom Count" in an O-Ring?

In general, 10.4x10^21 is considered a high "Gold Atom Count" for an O-Ring. The exact number may vary depending on the size and composition of the O-Ring, but this value indicates a significant amount of gold atoms present.

4. Can the "Gold Atom Count" in an O-Ring vary?

Yes, the "Gold Atom Count" in an O-Ring can vary depending on the manufacturing process and the materials used. Additionally, external factors such as exposure to certain chemicals or environments can also affect the number of gold atoms present in an O-Ring.

5. What are the potential applications of knowing the "Gold Atom Count" in an O-Ring?

Knowing the "Gold Atom Count" in an O-Ring can have various applications in scientific research. It can help in tracking the movement or behavior of the O-Ring in experiments, as well as provide insights into the composition and quality of the O-Ring. Additionally, this information can also be used in quality control and testing processes for O-Ring production.

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