Atom Volume Calc: Find the Volume of a Lead Atom

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    Atom Lead Volume
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The discussion revolves around calculating the volume of a lead atom, given its mass and density. The density of lead is 11.35 g/cm³, and the mass of a single lead atom is 3.439×10^-25 kg. Participants attempted to convert units and calculate the volume, with one user initially arriving at an incorrect result of 3.03×10^-26 m³. After further discussion, it was clarified that the correct volume should be 3.03×10^-29 m³, emphasizing the importance of accurate unit conversions. The conversation highlights the challenges of understanding atomic volume calculations in relation to density.
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Homework Statement


Lead has a mass of 11.35 g per cubic centimeter of volume, and the mass of one of its atoms is 3.439×10-25 kg. If the atoms are spherical and tightly packed, what is the volume of an individual atom? ( m3)

Homework Equations


conversions

The Attempt at a Solution


For this problem I converted the 11.35 g/cm ^3 into 11350 g/m^3 and then converted the atom from 3.439*10^-25 kg into 3.439*10^-22 g. Then I divided it by the 11350 to get m^3 and the result was 3.03*10^-26 but it appears to be wrong.
 
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Kpgabriel said:
spherical and tightly packed,
 
Kpgabriel said:

Homework Statement


Lead has a mass of 11.35 g per cubic centimeter of volume, and the mass of one of its atoms is 3.439×10-25 kg. If the atoms are spherical and tightly packed, what is the volume of an individual atom? ( m3)

Homework Equations


conversions

The Attempt at a Solution


For this problem I converted the 11.35 g/cm ^3 into 11350 g/m^3 and then converted the atom from 3.439*10^-25 kg into 3.439*10^-22 g. Then I divided it by the 11350 to get m^3 and the result was 3.03*10^-26 but it appears to be wrong.

How many ##cm## are there in a ##m##?
 
Last edited:
PeroK said:
How many ##cm## are there is a ##m##?
100 so it would be cubed to get m^3 and it would be 3.03*10^-29 which is the right answer
 
If 11,35 grams fit into 1 cm3 then 3.439×10-22 g will fitt into
3.439×10-22g/ 11.35= 3.03 cm3
 
Mathijsgri said:
If 11,35 grams fit into 1 cm3 then 3.439×10-22 g will fitt into
3.439×10-22g/ 11.35= 3.03 cm3
* 3.03*10-23 cm3
 
Mathijsgri said:
* 3.03*10-23 cm3
But you're looking for m^3
 
Kpgabriel said:
But you're looking for m^3
1m3= 1.000.000cm3
? = 3.03*10 - 23cm3
 
Kpgabriel said:
But you're looking for m^3

Are you helping him now? o_O
 
  • #10
'crosswise multiplication table' i don't know if it is an english term, i translated it literally from dutch.
 
  • #11
Mathijsgri said:
1m3= 1.000.000cm3
? = 3.03*10 - 23cm3
Right. so it would be 11.35 g/cm^3 * 1000000cm^3/1 m^3 to give 11350000 g/m^3 which you divide into 3.439*10^-22 g
 
  • #12
Mathijsgri said:
'crosswise multiplication table' i don't know if it is an english term, i translated it literally from dutch.
I just call it a conversion
 
  • #13
Kpgabriel said:
Right. so it would be 11.35 g/cm^3 * 1000000cm^3/1 m^3 to give 11350000 g/m^3 which you divide into 3.439*10^-22 g
keep in mind that 1 cm3 is not equalle to 1.000.000 m3
 
  • #14
Mathijsgri said:
keep in mind that 1 cm3 is not equalle to 1.000.000 m3

I don't think you're helping here. The OP has got the correct answer. Leaving aside the question of the basic volume of a sphere as opposed to the volume it occupies in a 3D array.
 
  • #15
Kpgabriel said:
100 so it would be cubed to get m^3 and it would be 3.03*10^-29 which is the right answer

I assume that's the answer on your answer sheet?
 
  • #16
you are right i made a typo in my calculator, sorry for that.
 
  • #17
PeroK said:
I assume that's the answer on your answer sheet?
Yes it was a test question I got wrong so I was trying to find the correct method to the answer
 

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