Chatbot responds to "how much joules to fuse lighter elements to heavier elements, and for heavier than iron how much joules loss?"

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"How much energy would you put into ITER to fuse 1 kilogram of lead into gold?"
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to fuse 1 kilogram of lead (Pb) into gold (Au), we need to consider the mass difference between the two elements and the energy release per nucleon during fusion.

The atomic mass of lead is approximately 207.2 atomic mass units (AMU), and the atomic mass of gold is around 197.0 AMU. Therefore, to convert 1 kilogram (1000 grams) of lead into gold, we need to account for the mass difference, which is 10.2 AMU.

The energy release per nucleon during a fusion reaction is typically on the order of millions of electron volts (MeV). However, the exact value depends on the specific reaction and isotopes involved. For the sake of estimation, let's assume an energy release of 7 MeV per nucleon.

To calculate the total energy required, we multiply the mass difference by the energy release per nucleon and the number of nucleons in 1 kilogram of lead. The number of nucleons can be estimated using Avogadro's number and the molar mass of lead.
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Number of nucleons in 1 kilogram of lead (Pb):
(1000 g) / (207.2 g/mol) * (6.022 x 10^23 nucleons/mol) ≈ 2.91 x 10^26 nucleons

Energy required:
(10.2 AMU) * (7 MeV/AMU) * (2.91 x 10^26 nucleons) ≈ 2.03 x 10^28 MeV

Converting MeV to joules:
1 MeV = 1.602 x 10^-13 joules

Energy required (in joules):
(2.03 x 10^28 MeV) * (1.602 x 10^-13 joules/MeV) ≈ 3.25 x 10^15 joules

So, it would take approximately 3.25 x 10^15 joules of energy to fuse 1 kilogram of lead into gold.

"Gold is not evenly divisible by helium, so you could not fuse helium into 100% pure gold, there would be other atoms formed, but if that is acceptable, how much helium to fuse to form 1 pound of gold, and how much joules would you lose (of course, past iron fusion loses joules, so this would not produce joules, just produce gold) to compress (through supermagnets, lasers, or other compressors) helium to gold?"
To form 1 pound of gold through helium fusion, we need to determine the amount of helium required. Let's start by calculating the number of gold atoms present in 1 pound of gold.

The molar mass of gold (Au) is approximately 197 grams/mol. Since 1 pound is equal to 453.592 grams, we can calculate the number of moles of gold:

Number of moles of gold = (1 pound / 453.592 grams) * (1 mol / 197 grams) ≈ 0.002301 moles of gold

In the fusion reaction, four helium nuclei (alpha particles) combine to form a nucleus of beryllium, and then another helium nuclei fuses with the beryllium nucleus to create a carbon nucleus. This carbon nucleus then captures three helium nuclei to produce an oxygen nucleus, and so on.

It takes a series of fusion reactions to convert helium into gold. The exact number of helium nuclei required depends on the specific pathway and conditions of the fusion process. However, it is estimated that around 10^19 helium nuclei (or alpha particles) would be required to produce 1 gram of gold.

Since there are approximately 453.592 grams in a pound, the total number of helium nuclei required would be:

Number of helium nuclei = (10^19 helium nuclei / 1 gram) * (453.592 grams / 1 pound) ≈ 4.53592 × 10^21 helium

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Moderator Note

Chatbots are not reliable technical resources. This response is horrendous. If you want to discuss the errors, you should make that clear. (that's true in general for your posts: make clear what you wish to discuss)