Problem 87 - University Physics with Modern Physics

Question

Nuclear Fission. The unstable nucleus of uranium- 236 can be regarded as a uniformly charged sphere of charge \(Q=+92 e\) and radius \(R=\) \(7.4 \times 10^{-15} \mathrm{m}\) . In nuclear fission, this can divide into twosion, this can divide into two smaller nuclei, each with half the charge and half the volume of the original uranium- 236 nucleus. This is one of the reactions that occurred in the nuclear weapon that exploded over Hiroshima, Japan, in August 1945 . (a) Find the radii of the two "daughter" nuclei of charge \(+46 e\) . (b) In a simple model for the fission process, immediately after the uranium- 236 nucleus has undergone fission, the "daughter" nuclei are at rest and just touching, as shown in Fig. 23.41 . Calculate the kinetic energy that each of the "daughter" nuclei will have when they are very far apart. (c) In this model the sum of the kinetic energics of the two "daugh-teer" nuclei, calculated in part \((b),\) is the energy released by the fission of one uranium- 236 nucleus. Calculate the energy released bythe fission of 10.0 kg of uraninm- 236 . The atomic mass of nranium-236 is \(236 \mathrm{u},\) where \(1 \mathrm{u}=1\) atomic mass unit \(=1.66 \times 10^{-24} \mathrm{kg}\) of this model, discuss why an atomic bomb could just as well be called an "electric bomb."

Step-by-Step Solution

Answer

(a) Radii: approximately \(5.9 \times 10^{-15} \ m\). (b) Kinetic energy of each nucleus: \(1.7 \times 10^{-11} J\). (c) Energy released from 10 kg: \(8.67 \times 10^{14} J\).

Step 1: Calculating the Radii of Daughter Nuclei

For part (a), the volume of the original nucleus is given by the formula for the volume of a sphere: \( V = \frac{4}{3} \pi R^3 \). Since each daughter nucleus has half the volume of the original, the volume of a daughter nucleus is \( V_d = \frac{1}{2} V = \frac{2}{3} \pi R^3 \). Each daughter nucleus is also a sphere, so the volume equation becomes \( \frac{4}{3} \pi r^3 = \frac{2}{3} \pi R^3 \). Solving for \( r \), we find:\[ r^3 = \frac{1}{2} R^3 \]\[ r = \left( \frac{1}{2} \right)^{1/3} R \]Substituting \( R = 7.4 \times 10^{-15} \ m \) we get:\[ r = \left( \frac{1}{2} \right)^{1/3} \times 7.4 \times 10^{-15} \ m \approx 5.9 \times 10^{-15} \ m \]

Step 2: Finding Potential Energy when Daughter Nuclei are Very Far Apart

For part (b), we calculate the electrostatic potential energy between the daughter nuclei when very far apart. Initially, when just touching, the distance between their centers is \( 2r \). Use the formula for potential energy of two point charges:\[ U = \frac{k \cdot Q_1 \cdot Q_2}{r} \]where \( k = 8.99 \times 10^9 \, N \cdot m^2/C^2 \), \( Q_1 = Q_2 = 46e \), and \( r = 2 \times 5.9 \times 10^{-15} \ m \).The charge \( e = 1.6 \times 10^{-19} \ C \), so the charges are \( Q_1 = Q_2 = 46 \times 1.6 \times 10^{-19} \ C \).Substituting into the formula, we find:\[ U = \frac{(8.99 \times 10^9) \cdot (46 \times 1.6 \times 10^{-19})^2}{2 \times 5.9 \times 10^{-15}} \approx 1.7 \times 10^{-11} \ J \]

Step 3: Calculating Energy Released from Fission of Uranium-236

For part (c), first calculate the energy released from one uranium-236 fission using the kinetic energy found in step 2, but note that this energy is doubled for both nuclei moving apart. Thus:\[ E_{total} = 2 \times U = 2 \times 1.7 \times 10^{-11} \ J = 3.4 \times 10^{-11} \ J \]Next, calculate the number of uranium nuclei in 10 kg using the atomic mass:\[ \text{Number of nuclei} = \frac{10 \ kg}{236 \ AMU} \times \frac{1}{1.66 \times 10^{-27} \ kg/AMU} \approx 2.55 \times 10^{25} \text{ nuclei} \]Finally, multiply by the energy per fission:\[ E_{released} = 2.55 \times 10^{25} \times 3.4 \times 10^{-11} \approx 8.67 \times 10^{14} \ J \]

Step 4: Explaining the Concept of an 'Electric Bomb'

The energy calculation shows that the majority of the energy released in fission is due to the conversion of electrostatic potential energy to kinetic energy as a result of nuclear forces overcoming the electric repulsion between protons. Thus, an atomic bomb could be considered an "electric bomb" in the sense that it releases stored electrostatic potential energy.

Key Concepts

Electrostatic Potential Energy

Electrostatic potential energy plays a significant role in nuclear reactions, especially in nuclear fission. It represents the energy stored due to the electrical interactions between charged particles within the nucleus—primarily protons. Protons are positively charged and repel each other due to electrostatic forces. Despite this repulsion, nuclear forces help hold the nucleus together.
In the case of nuclear fission, like with uranium-236, when the nucleus is split into smaller "daughter" nuclei, electrostatic repulsion becomes significant. As the daughter nuclei separate, the electrostatic repulsion between their protons increases, converting the stored potential energy into kinetic energy. This release of energy is what makes nuclear fission reactions highly energetic.
Understanding electrostatic potential energy is crucial for comprehending why nuclei fission results in such a massive energy release. This potential energy is transformed into kinetic energy, which contributes to the powerful explosions seen in nuclear bombs.

Kinetic Energy in Nuclear Reactions

When nuclear fission occurs, such as the splitting of uranium-236, the resultant "daughter" nuclei move apart with substantial speed. This motion is attributable to the kinetic energy they gain.
Kinetic energy in nuclear reactions refers to the energy of motion resulting from the conversion of electrostatic potential energy among the nucleus's protons. When the daughter nuclei are formed, they initially start at rest but, as they move apart due to Coulomb forces, they convert potential energy into kinetic energy. The formula for calculating kinetic energy can be seen in: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass, and \( v \) is the velocity of the nucleus.
In fission reactions, like in the original problem, calculating the kinetic energy helps in determining how much energy each nucleus will have when it is far apart. This also directly correlates to the energy simply released by the fission process. Massive amounts of energy are produced due to the rapid separation of the highly charged daughter nuclei.

Atomic Bomb Mechanism

Nuclear reactions, such as fission, are the basis of atomic bombs. These devices are sometimes referred to as "electric bombs" due to the central role of electrostatic potential energy. When a nuclear bomb detonates, it triggers a chain reaction. Each fission event releases neutrons, which trigger further fission reactions in nearby nuclei.
The primary mechanism involves the rapid release of energy stored as electrostatic potential energy in atomic nuclei. As these nuclei split, the potential energy is transformed primarily into kinetic energy of the fragments, with substantial energy emerging as heat and radiation.
The immense destructive power of an atomic bomb comes from the conversion of a small amount of matter into vast amounts of energy, as described by Einstein's famous equation \( E=mc^2 \). In practice, the detonation shakes the surrounding air, resulting in a massive, destructive blast. Understanding the "electric bomb" concept helps explain the explosive energy, grounded in the forces that propagate within the atomic structure.