If you started off with one higrigen atom and doubled it every second (1,2,4,8,16,ect…), how long would it take to fill up a space as big as the known universe. The hydrogen would be in gas form, but at a density of right before it turned liquid.

Bonus question. Start off with a neutron and start doubling it (in essence creating a neutron star). How long would that take?

Higriger is not listed in the periodic table.
You may need a “Quibic” quantum computer to solve.
More homework for dopers? :dubious:
Didn’t the instructor furnish the data?

First, it is hydrogen. Second, atomic mass does not equal volume. Third, there are many different radii I could use. I am going to use atomic radius for this problem. Fourth, hydrogen is diatomic. In nature you will not find one hydrogen atom by itself but rather two bonded together. With all that being said, here is my take on it.

Volume of known universe (assuming it is spherical): 1.9 × 10^33 cubic light years
1 light year is roughly equal to 9.46 trillion km
Atomic radius of one hydrogen atom: 25 picometers or 2.5 × 10^-14 km (this will help us later)
Equation of a circle: 4/3pir^3
The volume of a hydrogen atom would there for be 6.54498 x 10^-41 cubic km

We now want an exponential equation
ln(Y/X)=t*k

When t=2 we have ln(1.309 x 10^-40/6.54498 x 10^-41)=2*k
Solving for k we now have a constant of 0.34657359

You’d never be able to fill the universe to that density of anything, as it would collapse under its own gravitation long before the universe was anywhere near full. However, we can make a few assumptions and ignore physics completely. If we assume by “known universe” you mean the sphere enclosing the visible universe, that we’re dealing with a sphere with a radius of about 15 billion LY. Your requirement about the particle density is a bit vague. For clarity, we’ll use STP, at which (according to this page) that there are about 700 billion billion hydrogen atoms per cm[sup]3[/sup] of pure hydrogen gas. To fill up a sphere 15 billion LY in radius to this density would take 1.17 x 10[sup]78[/sup] atoms. Doubling once per second starting at one would take a mere 260 seconds! (2^260 = 1.85 x 10[sup]78[/sup]. Spingears: :rolleyes:

This is getting tiresome, spingears. If you think that someone is asking homework questions, we’ve devised a method for you to report this. It’s actually less time consuming than writing a snarky reply to the OP.

HIT THE REPORT BAD POST button in the upper right corner of the actual posts. It’s a ! symbol.

Yea, I did not even worry about it being in the form of a gas. What I could of done was taken the molar volume and changed it for when the substance is “right before [its] liquid form” but did not even mess with it. Oh well.

5 replies to one thread. I need to quit. Anyways, yes, you are right. Hydrogen exsists as a single atom in some situations simply because there is no way for it to combine with other hydrogen atoms. I, however, was applying a standard situation to the hydrogen atoms. I shouldn’t of, but I do not think it matters either way.

It is indeed hydrogen and not Higriger (I ran this through the spell checker. I don’t know how this happened).

As for this being a homework question, you are wrong. This is a question I thought up last night in bed right before I fell asleep (a time when most of my best questions come to me). Although I agree with the SDMB policy on not answering homework questions, I find your unsupported assumption slightly insulting.

Thanks to everyone who contributed to the thread. I would have thought it would have taken a lot longer. Wow, under 10 minutes. I was thinking days.

First, could we settle on the volume of the known Universe?
Assuming a Universe of radius 15 billion light years,
I get (4/3)PI1.5X10^10 Light Years
= 1.4137 * 10[sup]31[/sup] Light Years

Nothing is diatomic inside the core of stars. The heat required for a fusion reaction to take place is far far greater than what would be needed to tear apart any chemical bond.

All figures for the radius of the universe involve a bit of guesswork, and the numbers may change as more powerful telescopes are invented.

Try again. First of all, use the correct formula which is 4/3 x pi x r[sup]2[/sup]. And check your math. I get 9.42 x 10[sup]20[/sup] LY[sup]3[/sup]. I think you slipped a decimal place or ten someplace.

Anyways, I just realized I messed something ELSE up with my calculation. I am the nation’s official bumbling, awkward, and all around goofy novice mathematician.

Or 5.81666657 minutes. Yes, I did not do the right volume for Hydrogen gas right before it turns into a liquid. Hopefully I will have this finally resolved (my calculations atleast). Bastard me.

although my number came out correctly, I forgot to put the exponent in the calculation which should be:

I get (4/3)PI(1.5X10[sup]10[/sup])[sup]3[/sup] Light Years

Geez, volume calculations can be very tricky. I previewed my calculation, saw it already had ONE exponent but I forgot to show that raised to another exponent. You figure 1 exponent per number always looks right.

So, I guess we all are guilty of committing a few mathematical* faux pas * in this thread. :smack:

(I am SO glad my Ultar Converter was right though).