Okay, so I need to figure out the energy of the Relativistic Heavy Ion Collider (RHIC) such that the speed of the nucleus X is 0.96c. The nucleus X is given as Li with 3 neutrons. Wait, lithium normally has 3 protons and 3 neutrons, right? But the question says Li with 3 neutrons. So maybe it's Li-6? Because lithium-6 has 3 protons and 3 neutrons. Let me confirm that. Yes, lithium has two main isotopes: Li-6 and Li-7. Li-6 has 3 protons and 3 neutrons, so atomic number 3, mass number 6. So the nucleus X here is Li-6. Got it.

The problem is asking for the energy of RHIC such that the nucleus is moving at 0.96c. Since RHIC is a heavy ion collider, it's likely dealing with ions, which are nuclei of atoms. So here, the nucleus is Li-6. We need to compute the energy required to accelerate this nucleus to 0.96c.

Hmm. Energy in relativity. For relativistic speeds, we can't use the classical kinetic energy formula. Instead, we need to use the relativistic kinetic energy formula. The total energy of a particle is given by gamma times m0c², where gamma is the Lorentz factor. Then the kinetic energy is the total energy minus the rest energy, so (gamma - 1)m0c².

So the steps would be:

1. Find the rest mass of the nucleus (Li-6).
2. Calculate gamma for v = 0.96c.
3. Compute the kinetic energy (gamma - 1)m0c².
4. Convert the energy into GeV.

But wait, the problem mentions the energy of the collider. Since RHIC collides heavy ions, the energy per nucleon is a common measure. Wait, but the question is about the energy of the collider such that the nucleus (the entire ion) reaches 0.96c. That seems a bit odd because RHIC typically uses different beams for different ions, and the energy is usually given per nucleon or per beam. But the question here is about the total energy required for the entire nucleus to reach that speed. Let me think.

Wait, maybe I need to clarify. If RHIC is colliding two ions head-on, each with some energy, then the total energy in the center-of-momentum frame would be higher. But the problem states "the speed of the nucleus X is equal to 0.96c". So perhaps it's a single ion being accelerated to 0.96c, not the center-of-mass frame. But RHIC is a collider, so it's more likely about colliding two ions. However, the question is phrased as "the energy of the Relativistic Heavy Ion Collider (RHIC) so that the speed of the nucleus X is equal to 0.96c". Wait, maybe it's not a collider but a fixed target experiment? But RHIC is a collider. Hmm. Alternatively, maybe the question is considering each beam's energy contributing to the speed of the nucleus. But I'm a bit confused here. Let me read the question again.

"What is the correct answer to this question: What is the energy of the Relativistic Heavy Ion Collider (RHIC) so that the speed of the nucleus X is equal to 0.96c?"

So the energy of RHIC is such that the nucleus X (Li-6) moves at 0.96c. Wait, perhaps RHIC accelerates the nucleus to that speed, not the beam. But RHIC typically accelerates protons or ions in each beam. If it's a heavy ion collider, the ions are being accelerated. But the problem is about the nucleus X (a Li-6 ion) moving at 0.96c. So maybe the question is assuming that each nucleus in the beam is moving at 0.96c, and we need to find the energy required per nucleus. But RHIC's energy is usually stated per nucleon or per beam. Hmm. Alternatively, perhaps the question is considering the entire ion as a single particle, but that's unusual. Maybe the problem is simplified, treating the entire nucleus as a single particle with mass equal to the rest mass of the nucleus. Let me proceed under that assumption for now.

So first, find the mass of Li-6. Lithium has an atomic number of 3 (Z=3), so the number of protons is 3. The mass number is 6, so the number of neutrons is 3. The rest mass of a proton is approximately 938.272 MeV/c², and a neutron is about 939.565 MeV/c². So the mass of Li-6 would be 3*(938.272) + 3*(939.565) = 3*938.272 + 3*939.565. Let me compute that:

3*938.272 = 2814.816 MeV/c²

3*939.565 = 2818.695 MeV/c²

Total = 2814.816 + 2818.695 = 5633.511 MeV/c². So the rest mass energy is 5633.511 MeV, which is 5.633511 GeV. Wait, but wait, wait. Wait, protons and neutrons have slightly different masses. Let me check the exact numbers. The exact mass of a proton is approximately 938.272 MeV/c², and a neutron is approximately 939.565 MeV/c². So yes, those numbers. So Li-6 has 3 protons and 3 neutrons. So total mass is 3*(938.272 + 939.565) = 3*(1877.837) = 5633.511 MeV/c². So 5633.511 MeV is 5.633511 GeV. So m0c² = 5.633511 GeV.

Now, the speed is 0.96c. So gamma = 1 / sqrt(1 - v²/c²) = 1 / sqrt(1 - (0.96)^2) = 1 / sqrt(1 - 0.9216) = 1 / sqrt(0.0784) = 1 / 0.28 = approximately 3.5714286. Wait, 0.0784 is 0.28 squared. Let me check: 0.28 * 0.28 = 0.0784. Yes. So gamma = 1/0.28 ≈ 3.5714286.

So gamma is approximately 3.5714286. Then the kinetic energy is (gamma - 1) * m0c². So (3.5714286 - 1) * 5.633511 GeV = 2.5714286 * 5.633511 GeV. Let's compute that.

First, 2.5714286 * 5 = 12.857143

2.5714286 * 0.633511 ≈ 2.5714286 * 0.6 = 1.542857 + 2.5714286 * 0.033511 ≈ 1.542857 + 0.08613 ≈ 1.629. So total ≈ 12.857143 + 1.629 ≈ 14.486 GeV. Wait, but let me do it more accurately.

Alternatively, 2.5714286 * 5.633511. Let's compute 2.5714286 * 5 = 12.857143, 2.5714286 * 0.633511. Let's compute 0.633511 * 2.5714286:

First, 0.6 * 2.5714286 = 1.542857

0.033511 * 2.5714286 ≈ 0.08613 (approx)

So total ≈ 1.542857 + 0.08613 ≈ 1.628987

So total kinetic energy ≈ 12.857143 + 1.628987 ≈ 14.48613 GeV. So approximately 14.486 GeV. But wait, the answer options are A:20.132, B:23.069, C:18.475, D:21.419. None of these are close to 14.48. So this suggests that my approach is wrong. Hmm. Maybe I made a mistake in the assumption.

Wait, perhaps the problem is considering the nucleus as a single particle, but in reality, heavy ions are multi-nucleon systems. However, in high-energy physics, when dealing with the kinetic energy of a nucleus, sometimes the concept of "nuclear rest mass" is used, but the way energy is calculated might be different. Alternatively, maybe the problem is considering the entire nucleus as a single particle, but the energy required is the total energy (rest plus kinetic) rather than just the kinetic energy. Wait, the question says "the energy of the Relativistic Heavy Ion Collider (RHIC) so that the speed of the nucleus X is equal to 0.96c". So perhaps it's the total energy of the nucleus, not the kinetic energy. So total energy is gamma * m0c². Then the energy would be gamma * 5.633511 GeV ≈ 3.5714286 * 5.633511 ≈ let's compute that.

3.5714286 * 5.633511. Let's compute 3 * 5.633511 = 16.900533, 0.5714286 * 5.633511 ≈ 0.5 * 5.633511 = 2.8167555, 0.0714286 * 5.633511 ≈ 0.4046. So total ≈ 2.8167555 + 0.4046 ≈ 3.2213555. So total energy ≈ 16.900533 + 3.2213555 ≈ 20.1218885 GeV. So approximately 20.122 GeV. Looking at the options, A is 20.132 GeV. That's very close. So maybe the answer is A. But wait, the problem says "the energy of the Relativistic Heavy Ion Collider (RHIC) so that the speed of the nucleus X is equal to 0.96c". If RHIC is providing that total energy to the nucleus, then 20.122 GeV is close to option A. But let me check the calculation again.

Wait, gamma = 1 / sqrt(1 - v²/c²) = 1 / sqrt(1 - 0.96²) = 1 / sqrt(0.0784) = 1/0.28 = 3.5714285714. So gamma is exactly 3.5714285714. So m0c² is 5.633511 GeV. So gamma * m0c² = 3.5714285714 * 5.633511. Let me compute that more accurately.

First, 3 * 5.633511 = 16.900533

0.5714285714 * 5.633511:

0.5 * 5.633511 = 2.8167555

0.0714285714 * 5.633511 ≈ 5.633511 * (1/14) ≈ 0.40239364

So total 2.8167555 + 0.40239364 ≈ 3.21914914

So total gamma * m0c² ≈ 16.900533 + 3.21914914 ≈ 20.11968214 GeV. Approximately 20.1197 GeV. The options are A:20.132, which is 20.132 GeV. So the difference is about 0.0123 GeV, which is about 12.3 MeV. The problem states that the precision is at 1e-4, so maybe the exact calculation would give a value close to 20.132 GeV. Let me check with more precise numbers.

Wait, perhaps the mass calculation is off. Let me recalculate the mass of Li-6 more accurately. The exact mass of a proton is approximately 938.272 MeV/c², and a neutron is approximately 939.565 MeV/c². So Li-6 has 3 protons and 3 neutrons. So total mass is:

3 * 938.272 + 3 * 939.565 = let's compute each:

3 * 938.272 = 2814.816 MeV/c²

3 * 939.565 = 2818.695 MeV/c²

Total = 2814.816 + 2818.695 = 5633.511 MeV/c². So that's correct. So m0c² is 5633.511 MeV, which is exactly 5.633511 GeV.

Gamma is exactly 1 / sqrt(1 - 0.96²) = 1 / sqrt(0.0784) = 1/0.28 = 3.5714285714285714...

So gamma * m0c² = 3.5714285714285714 * 5.633511 GeV. Let's compute this precisely.

3.5714285714 * 5.633511:

First, 3 * 5.633511 = 16.900533

0.5714285714 * 5.633511:

0.5 * 5.633511 = 2.8167555

0.0714285714 * 5.633511: Let's compute 5.633511 * 0.0714285714.

0.0714285714 is 1/14. So 5.633511 / 14 ≈ 0.40239364

So total 2.8167555 + 0.40239364 = 3.21914914

Total gamma * m0c² = 16.900533 + 3.21914914 = 20.11968214 GeV. So approximately 20.1197 GeV, which rounds to 20.120 GeV. The closest option is A:20.132 GeV. The difference is about 0.0123 GeV, which is 12.3 MeV. That's a small discrepancy. Maybe the problem uses slightly different mass values for protons and neutrons. Let me check the exact masses.

Looking up the exact masses:

Proton mass: 938.272 MeV/c²

Neutron mass: 939.565 MeV/c²

So Li-6 has 3 protons and 3 neutrons. So total = 3*938.272 + 3*939.565 = 2814.816 + 2818.695 = 5633.511 MeV/c². So that's correct.

Alternatively, maybe the problem considers the nucleus as a single particle and uses the rest mass of the nucleus, but maybe there's a factor I'm missing. Wait, but in relativity, the total energy is gamma*m0c², which is the total energy of the particle. But if the question is asking for the energy of the collider such that the nucleus moves at 0.96c, then the total energy provided by the collider to the nucleus would be gamma*m0c². So that's what we calculated, approximately 20.12 GeV. The closest option is A:20.132 GeV. But maybe the exact calculation with more precise gamma and m0c² would give a slightly higher value. Let me check with more precise numbers.

Let me compute gamma more precisely. v = 0.96c. So v²/c² = 0.9216. 1 - 0.9216 = 0.0784. sqrt(0.0784) = 0.28. So gamma = 1/0.28 = 3.5714285714285714... So exact gamma is 3.5714285714285714...

Then 3.5714285714285714 * 5.633511 GeV.

Let me compute this multiplication precisely.

3.5714285714 * 5.633511:

First, multiply 3 * 5.633511 = 16.900533

Then, 0.5714285714 * 5.633511:

0.5 * 5.633511 = 2.8167555

0.0714285714 * 5.633511:

Multiply 5.633511 by 0.0714285714. Let's compute this as:

0.0714285714 is 1/14. So 5.633511 / 14 = 0.402393642857

So total 0.5714285714 * 5.633511 = 2.8167555 + 0.402393642857 = 3.219149142857

So total gamma * m0c² = 16.900533 + 3.219149142857 = 20.119682142857 GeV ≈ 20.11968 GeV. So approximately 20.1197 GeV. The options are A:20.132, which is 0.0123 GeV higher. Hmm. Maybe the problem uses different values for the masses. Let me check the mass of Li-6 again.

Wait, maybe the problem refers to lithium-7 instead of lithium-6? Wait, the question says "Li with 3 neutrons". Lithium has an atomic number of 3. So if it has 3 neutrons, then the mass number is 3 + 3 = 6. So Li-6. So that's correct. If it were Li-7, it would have 4 neutrons (3 protons + 4 neutrons = 7), but the question says 3 neutrons. So Li-6 is correct.

Alternatively, maybe the problem uses a different value for the speed. Wait, 0.96c. Let me check if the calculation is correct. 0.96 squared is 0.9216, 1 - 0.9216 = 0.0784, square root is 0.28, so gamma is 1/0.28 = 3.5714. That's correct.

Wait, perhaps the problem is considering the nucleus as a composite particle, so the mass should be calculated differently? For example, if the nucleus is considered as a nucleus with binding energy, but that's probably not relevant here. The problem states "the energy of the Relativistic Heavy Ion Collider (RHIC) so that the speed of the nucleus X is equal to 0.96c". So the entire nucleus is moving at 0.96c, so the total energy required is gamma * m0c², which we calculated as approximately 20.1197 GeV. The closest option is A:20.132 GeV. The difference is about 0.0123 GeV, which is about a 0.06% difference. That's within the 1e-4 precision mentioned. So perhaps the answer is A.

Alternatively, maybe there's a miscalculation in the mass. Let me check the mass calculation again. 3 protons at 938.272 MeV: 3 * 938.272 = 2814.816 MeV. 3 neutrons at 939.565 MeV: 3 * 939.565 = 2818.695 MeV. Total = 2814.816 + 2818.695 = 5633.511 MeV. So 5633.511 MeV is 5.633511 GeV. Correct.

Alternatively, maybe the problem uses a different value for the mass of the nucleus. Wait, sometimes the mass of the nucleus is approximated as the sum of the masses of the protons and neutrons minus the binding energy. But binding energy is usually in the order of MeV, so for a nucleus with 6 nucleons (3 protons, 3 neutrons), the binding energy would be small compared to the total mass. Let's check. The mass of a proton is about 938.272 MeV/c², neutron 939.565 MeV/c². The mass of a nucleus is usually calculated as Z*m_p + N*m_n - B, where B is the binding energy. But B is typically on the order of several MeV per nucleus. For example, for Li-6, the mass is 5633.511 MeV/c² minus the binding energy. Wait, but the question is giving the nucleus as Li with 3 neutrons. So Li-6. Let me check the actual mass of Li-6. According to the periodic table, lithium-6 has a molar mass of approximately 6.015 g/mol. But the exact atomic mass is 6.015 g/mol, which corresponds to a mass of 6.015 u (atomic mass units). 1 u is approximately 931.5 MeV/c². So 6.015 u * 931.5 MeV/u ≈ 5600 MeV. But our calculation was 5633.511 MeV. There's a discrepancy here. Wait, maybe the exact mass of Li-6 is lower. Let me check the exact mass. According to data, the exact atomic mass of lithium-6 is approximately 6.015 u. So 6.015 * 931.5 ≈ 6.015 * 931.5 = let's compute that.

6 * 931.5 = 5589 MeV

0.015 * 931.5 = 13.9725 MeV

Total ≈ 5589 + 13.9725 ≈ 5602.9725 MeV. So approximately 5603 MeV. But our previous calculation was 5633.511 MeV. So there's a difference here. Which one is correct? The problem states that X is defined as Li with 3 neutrons, so that should be Li-6. The problem gives X as Li with 3 neutrons, so Z=3 (Li), N=3 (since 3 protons, 3 neutrons). So the nucleus is Li-6. The exact mass of Li-6 is approximately 6.015 u, which is about 5603 MeV. But according to our previous calculation, the sum of individual nucleons is 5633.511 MeV. So there's a discrepancy of about 30 MeV. This is due to the binding energy. The actual mass of the nucleus is the sum of the masses of the individual nucleons minus the binding energy. So the binding energy would be the difference between the sum of individual masses and the actual mass. So 5633.511 - 5603 ≈ 30.511 MeV. So the binding energy is about 30.511 MeV. Therefore, the mass of the nucleus is 5603 MeV/c². But in our previous calculation, we used the sum of individual masses (5633.511 MeV/c²) as the rest mass. However, the actual rest mass is 5603 MeV/c². So this would affect the calculation. Let's see. Using the actual mass of Li-6 (5603 MeV/c²), then gamma * m0c² = 3.5714285714 * 5603 MeV. Let's compute that.

3.5714285714 * 5603 = ?

First, 3 * 5603 = 16809

0.5714285714 * 5603:

0.5 * 5603 = 2801.5

0.0714285714 * 5603 ≈ 0.0714285714 * 5603 ≈ 400.000 (since 0.0714285714 is 1/14, 5603/14 ≈ 400.214)

So total 2801.5 + 400.214 ≈ 3201.714

Total gamma * m0c² ≈ 16809 + 3201.714 ≈ 20010.714 MeV, which is 20.0107 GeV. That's approximately 20.0107 GeV. The closest option is A:20.132 GeV, which is still about 0.121 GeV higher. So this suggests that using the actual nuclear mass (considering binding energy) gives a result closer to option A but not exactly matching. However, the problem states that X is defined as Li with 3 neutrons, but in reality, the actual rest mass of Li-6 is lower. So perhaps the problem expects us to use the sum of the individual nucleon masses (5633.511 MeV/c²) as the rest mass, even though in reality the binding energy would reduce that. Alternatively, maybe the problem is using the atomic mass of lithium, which is around 6.941 GeV/c². Wait, no. The atomic mass of lithium (Li) is about 6.941 GeV/c², but that's the atom, not the nucleus. The nucleus of Li-6 has a mass of approximately 5603 MeV/c². But the problem defines X as Li with 3 neutrons, which is Li-6, but if using the atomic mass (which includes electrons), that's different. However, the problem likely refers to the nuclear mass, so 5603 MeV/c². But this complicates things.

Alternatively, perhaps the problem is assuming that the nucleus is treated as a free particle and uses the sum of the masses of the protons and neutrons. The problem states that the nucleus X is Li with 3 neutrons, which is Li-6. So I think the intended approach is to use the sum of the individual nucleon masses (5633.511 MeV/c²) as the rest mass. Then the calculation would be gamma * m0c² ≈ 20.1197 GeV, which is closest to option A:20.132 GeV. The difference is about 0.0123 GeV, which is 12.3 MeV. Since the problem allows for a precision of 1e-4 (0.0001), but the difference here is about 0.0123 GeV, which is more than an order of magnitude higher than 1e-4. So perhaps there's a mistake in the problem's options or in the intended approach. Alternatively, maybe the problem is considering the nucleus as a single particle with a different mass.

Wait, another thought: maybe the problem is considering the entire nucleus as a free particle, but in reality, nuclei are made of protons and neutrons. So if the nucleus is moving at 0.96c, maybe the individual nucleons are moving at a different speed? But the problem states the speed of the nucleus, so the entire nucleus is moving at 0.96c. Therefore, the entire nucleus has that speed. So the total energy is gamma * m0c². Therefore, the calculation is correct as 20.1197 GeV. The closest option is A:20.132 GeV. But the discrepancy is about 0.0123 GeV. Maybe the problem expects us to use a different gamma value? Let me check the gamma calculation again. v = 0.96c. So gamma = 1 / sqrt(1 - 0.96²) = 1 / sqrt(1 - 0.9216) = 1 / sqrt(0.0784) = 1 / 0.28 = 3.5714285714. Correct. So gamma is precise. The mass calculation is 5633.511 MeV/c². So the product is 20.1197 GeV. So the answer should be approximately 20.120 GeV, which is closest to option A:20.132 GeV. The difference is about 0.0123 GeV. But perhaps the problem expects us to use more precise values for the masses. Let me check the exact mass of Li-6. According to precise data, the isotopic mass of lithium-6 is 56.0151 u. 1 u = 931.5 MeV/c². So 56.0151 u * 931.5 MeV/u ≈ 56.0151 * 931.5 ≈ let's compute that.

56 * 931.5 = 52,164 MeV

0.0151 * 931.5 ≈ 14.08965 MeV

Total ≈ 52,164 + 14.08965 ≈ 52,178.08965 MeV. So approximately 52,178.09 MeV. Wait, that can't be right. Wait, 56.0151 u is the atomic mass. Wait, no, lithium-6 has a mass number of 6, so the mass is 6.0151 u. So 6.0151 u * 931.5 MeV/u ≈ 6.0151 * 931.5 ≈ 5602.97 MeV. So approximately 5603 MeV. So the rest mass is 5603 MeV/c². So using that, gamma * m0c² = 3.5714285714 * 5603 ≈ 20010.714 MeV ≈ 20.0107 GeV. That's approximately 20.011 GeV. The closest option is A:20.132 GeV. The difference is about 0.121 GeV. That's still a significant discrepancy. So perhaps the problem is using the mass of a proton and neutron as individual particles and summing them, which leads to a higher mass. But in reality, the nucleus's mass is lower due to binding energy. However, the problem states that X is defined as Li with 3 neutrons, so perhaps the intended approach is to use the sum of individual nucleon masses. In that case, the answer would be approximately 20.1197 GeV, which rounds to 20.120 GeV, closest to option A:20.132 GeV. Given that the problem allows for a precision of 1e-4, perhaps the intended answer is A. Alternatively, maybe there's a mistake in the problem's options, and the correct answer is not listed. But given the options, the closest is A.

Alternatively, perhaps the problem is considering the nucleus as a free particle, but the energy required is the total energy (including rest mass) rather than kinetic. Wait, no. The question says "the energy of the Relativistic Heavy Ion Collider (RHIC) so that the speed of the nucleus X is equal to 0.96c". The energy required would be the total energy, which is gamma * m0c². So that's correct. Therefore, the answer is approximately 20.120 GeV, which matches option A:20.132 GeV when rounded to four decimal places. However, there's a slight discrepancy. Alternatively, maybe the problem uses a different value for the speed. Wait, perhaps the speed is given as 0.96c, but relativistic energy is calculated using the relativistic momentum formula. Wait, no. Relativistic momentum is gamma * m0v, but the energy is gamma * m0c². So the calculation is correct. Therefore, I think the intended answer is A:20.132 GeV.
</think>