What is the Average Density of the Universe and Its Scale Over Time?

[Thomas Smith]

Homework Statement

The current average density of the universe is roughly 3 x 10^-27 km m^-3. What was the average density of the universe at the time the light was emitted by a galaxy with the red shift of z=11.1? Express your answer in terms of a number of H atoms per cubic meter.

Relevant Equations

R(t)/R(t0) = 1/1+z

Na x p/mH Where Na is Avogardo's number in terms of atoms, mH is the mass of the hydrogen atom in kg and p is the average density.

Firstly i worked out the scale factor of the universe
R(t)/R(t0) = 1/1+z = 1/1+11.1 = 1/12.1 = 12.1^3 = 1/1772

The distance between the galaxies were 12.1 times less than today and the volume was 1772 times smaller than today.

Then I think the average density in the universe at that time is (3×10^-27 )×1772 = 5.32 × 10^-24kg m^-3

then the average density in terms of the hydrogen atom
= 6.023×10^23 × (5.32×10^-24/1.67×10^-24) = 1.92×10^24 hydrogen atoms per cubic meter.

This does not seem right to me at all!

 

[phyzguy]

If you have the density in kg/m^3, and you want the density in hydrogen atoms/m^3, then you need to divide by kg/hydrogen atom, which you correctly have as 1.6x10^-24. Why did you then multiply by Avogadro's number?

 

[Thomas Smith]

If you have the density in kg/m^3, and you want the density in hydrogen atoms/m^3, then you need to divide by kg/hydrogen atom, which you correctly have as 1.6x10^-24. Why did you then multiply by Avogadro's number?

I thought that in regards to the number of hydrogen atoms you needed to include it

 

[jbriggs444]

I thought that in regards to the number of hydrogen atoms you needed to include it

Including it is fine. Including it twice is not. If you had tracked units carefully, the error would have made made itself more obvious.

There is another error in your calculation. One that tracking units will not catch. If you scale down distance by a factor of 1772, what does that do to volume?

 

[phyzguy]

There is another error in your calculation. One that tracking units will not catch. If you scale down distance by a factor of 1772, what does that do to volume?

Maybe I'm wrong, but I think the OP did this correctly. The distance scales by a factor of 12.1, which scales the volume by 12.1^3 = 1772.

 

[phyzguy]

I thought that in regards to the number of hydrogen atoms you needed to include it

No. including it gives you the mass of a mole of hydrogen atoms. You want the mass of a single hydrogen atom.