A 66 kiloton atomic bomb is fuelled with pure
235
U, 4.0% of which actually undergoes
fission. [i] How much uranium is in the bomb? [ii] How many fission fragments are
produced? [iii] How many neutrons generated in the fissions are released to the
environment? (On an average, each fission of
235
U produces 2.5 neutrons. A bomb’s
rating is the magnitude of the released energy, specified in terms of the mass of TNT
required to produce the same energy release. One megaton [10
6
tons] of TNT releases
2.6 × 10
28
MeV of energy.)

To tackle your question about the 66 kiloton atomic bomb fueled with pure uranium-235 (U-235), we can break it down into three parts: calculating the amount of uranium in the bomb, determining the number of fission fragments produced, and figuring out how many neutrons are released into the environment. Let’s go through each part step by step.

Calculating the Amount of Uranium in the Bomb

The bomb's energy yield is given as 66 kilotons of TNT. First, we need to convert this energy into a more usable form. One kiloton of TNT is equivalent to approximately 4.184 × 10¹² joules. Therefore, 66 kilotons can be calculated as follows:

• Energy in joules = 66 × 4.184 × 10¹² J = 2.76 × 10¹⁴ J.

Next, we know that the energy released per fission of U-235 is about 200 MeV. To convert this to joules, we use the conversion factor where 1 MeV = 1.602 × 10^-13 joules:

• Energy per fission = 200 MeV × 1.602 × 10^-13 J/MeV = 3.204 × 10^-11 J.

Now, we can find the total number of fissions required to produce the energy of the bomb:

• Number of fissions = Total energy / Energy per fission = (2.76 × 10¹⁴ J) / (3.204 × 10^-11 J) ≈ 8.61 × 10²⁴ fissions.

Since only 4.0% of the U-235 undergoes fission, we can find the total amount of U-235 needed:

• Total fissions = 0.04 × Total U-235 atoms.
• Total U-235 atoms = (8.61 × 10²⁴ fissions) / 0.04 = 2.15 × 10²⁶ atoms.

To find the mass of U-235, we use Avogadro's number (approximately 6.022 × 10²³ atoms/mol) and the molar mass of U-235 (about 235 g/mol):

• Mass of U-235 = (2.15 × 10²⁶ atoms) / (6.022 × 10²³ atoms/mol) × 235 g/mol ≈ 8.43 × 10³ g = 8.43 kg.

Determining the Number of Fission Fragments Produced

Each fission of U-235 produces two or three fission fragments. For simplicity, let's assume an average of 2.5 fission fragments per fission:

• Total fission fragments = Number of fissions × Average fission fragments per fission = (8.61 × 10²⁴) × 2.5 ≈ 2.15 × 10²⁵ fission fragments.

Calculating Neutrons Released to the Environment

Each fission of U-235 also releases about 2.5 neutrons. Therefore, the total number of neutrons produced can be calculated as follows:

• Total neutrons = Number of fissions × Neutrons per fission = (8.61 × 10²⁴) × 2.5 ≈ 2.15 × 10²⁵ neutrons.

Since we are interested in the neutrons released to the environment, we can assume that a significant portion of these neutrons escapes during the explosion, although some may be absorbed in the bomb's material. However, for this calculation, we can consider that most of them are released.

Summary of Results

To summarize:

• Amount of Uranium in the Bomb: Approximately 8.43 kg of U-235.
• Number of Fission Fragments Produced: About 2.15 × 10²⁵ fission fragments.
• Neutrons Released to the Environment: Approximately 2.15 × 10²⁵ neutrons.

This breakdown illustrates the calculations involved in understanding the mechanics of a nuclear bomb fueled by U-235. Each step builds on the previous one, leading to a comprehensive understanding of the bomb's composition and the reactions occurring within it.

To tackle your question about the atomic bomb fueled with pure uranium-235, we can break it down into three parts: calculating the total amount of uranium in the bomb, determining the number of fission fragments produced, and figuring out how many neutrons are released into the environment. Let's dive into each part step by step.

Calculating the Amount of Uranium in the Bomb

The bomb has a yield of 66 kilotons. First, we need to convert this yield into energy. Since 1 kiloton of TNT is equivalent to approximately 4.184 × 10¹² joules, we can calculate the total energy released by the bomb:

• 66 kilotons = 66 × 4.184 × 10¹² joules = 2.76 × 10¹⁴ joules.

Next, we know that 1 megaton of TNT releases 2.6 × 10²⁸ MeV of energy. To convert joules to MeV, we use the conversion factor: 1 joule = 6.242 × 10¹² MeV. Thus, we can find the energy in MeV:

• Energy in MeV = (2.76 × 10¹⁴ joules) × (6.242 × 10¹² MeV/joule) ≈ 1.72 × 10²⁷ MeV.

Now, since only 4.0% of the uranium undergoes fission, we can find the total energy produced by the fission of uranium-235:

• Let E_fission be the energy released per fission of uranium-235, which is approximately 200 MeV.
• Let N be the number of fissions. The total energy from fission is then N × E_fission.
• Setting this equal to the energy released by the bomb gives us: N × 200 MeV = 1.72 × 10²⁷ MeV.
• Solving for N: N = (1.72 × 10²⁷ MeV) / (200 MeV) = 8.6 × 10²⁴ fissions.

Now, since each fission of uranium-235 consumes one atom of uranium, we need to find the total mass of uranium that corresponds to this number of fissions. The number of atoms in a mole (Avogadro's number) is approximately 6.022 × 10²³ atoms/mole. Thus, we can calculate the moles of uranium-235:

• Moles of uranium-235 = N / Avogadro's number = (8.6 × 10²⁴ fissions) / (6.022 × 10²³ atoms/mole) ≈ 14.3 moles.

Now, the molar mass of uranium-235 is about 235 g/mole. Therefore, the total mass of uranium-235 in the bomb is:

• Mass of uranium-235 = 14.3 moles × 235 g/mole ≈ 3360.5 g or approximately 3.36 kg.

Determining the Number of Fission Fragments Produced

Fission of uranium-235 typically produces two fission fragments per fission event. Therefore, the total number of fission fragments can be calculated as follows:

• Fission fragments = 2 × N = 2 × (8.6 × 10²⁴) = 1.72 × 10²⁵ fragments.

Calculating Neutrons Released to the Environment

Each fission event of uranium-235 releases, on average, about 2.5 neutrons. To find the total number of neutrons released, we can multiply the number of fissions by the average number of neutrons per fission:

• Total neutrons = N × 2.5 = (8.6 × 10²⁴) × 2.5 = 2.15 × 10²⁵ neutrons.

In summary, we have:

• Amount of uranium in the bomb: Approximately 3.36 kg of uranium-235.
• Number of fission fragments produced: About 1.72 × 10²⁵ fragments.
• Neutrons released to the environment: Approximately 2.15 × 10²⁵ neutrons.

This breakdown illustrates the calculations and reasoning behind each part of your question, providing a comprehensive understanding of the processes involved in the fission of uranium-235 in an atomic bomb.