Anyone good at chemistry????

Wow I have so much respect for you guys.

 

i'm bad ass with chemistry

you put your left foot in and you shake it all about

 

hahaha nice.

 

The jist of this problem is to get an idea of how many suns would have to be inside the volume of the earth if the earth was a neutron-type star (extremely dense).

I would put all values in mks units (meters, kg, seconds).

1. One can start by calculating the volume of the earth.
2. Calculate how many nuclei can fit in an earth volume.
3. Calculate the total mass of nuclei in an earth volume.
4. Calculate how many sun masses are needed to achieve this mass.

One Angstrom is 1E-10 meters.

d1 = diameter of lead nucleus = 7.8E-5 A = 7.8E-15 m

m1 = mass of lead nucleus = 207 amu = 207 amu * (1.66E-27 kg/amu) = 3.4E-25 kg

d2 = diameter of earth = 7826 mi * (5280 ft/mi) * (1 m / 3.281 ft) = 1.3E7 m

V2 = volume of earth = (4/3)(pi)*(d2/2)^3 = 1.05E21 m^3

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A decision needs to be made as to how to pack the spherical nuclei. I will assume a simple cubic close packing, meaning that there is one nucleus at each of the 8 corners of a cube. Each corner contains 1/8th of a nucleus. That means each cube contains one nucleus.

The width of a cube of nuclei is the diameter of the nucleus (you can see this from the geometry of a simple cubic packing arrangement)

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number of nuclei per unit volume = 1 nucleus / (d1)^3 = 1 / (7.8E-15 m)^3 = 2.1E42 nuclei / m^3

number of nuclei that can fit in earth = V2 * nuclei per unit volume = 1.05E21 m^3 * 2.1E42 nuclei / m^3 = 2.2E63

total mass of nuclei in earth volume = 2.2E63 * m1 = 2.2E63 * 3.4E-25 kg = 7.6E38 kg

The mass of a sun is 2E20 kg

Number of suns needed = 7.6E38 kg / 2E20 kg = 3.8E18 suns

That's a lot of suns.

.

 

Here's a good site for unit conversion. Just put in any two units and it will tell you the relation.

http://www.calchemy.com/uclive.htm

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Somehow all of this makes perfect sense to me


H