Problem 53 - University Physics with Modern Physics

Question

If two deuterium nuclei (charge \(+e\), mass 3.34 \(\times\) 10\(^{-27}\) kg) get close enough together, the attraction of the strong nuclear force will fuse them to make an isotope of helium, releasing vast amounts of energy. The range of this force is about 10\(^{-15}\) m. This is the principle behind the fusion reactor. The deuterium nuclei are moving much too fast to be contained by physical walls, so they are confined magnetically. (a) How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse? (Assume their speeds are equal. Treat the nuclei as point charges, and assume that a separation of 1.0 \(\times\) 10\(^{-15}\) is required for fusion.) (b) What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.50 m?

Step-by-Step Solution

Answer

Velocity: \(9.63 \times 10^6 \, \text{m/s}\), Magnetic field: \(1.60 \, \text{T}\).

Step 1: Calculate the Kinetic Energy Needed for Fusion

To cause fusion, the kinetic energy (KE) of each nucleus must equal the Coulomb potential energy (PE) at a separation of \(1.0 \times 10^{-15}\) m. The Coulomb potential energy is given by the formula: \[PE = \frac{k \cdot e^2}{r}\]where \(k\) is the Coulomb's constant \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\), \(e\) is the elementary charge \(1.6 \times 10^{-19}\) C, and \(r\) is the separation distance (\(1.0 \times 10^{-15}\) m). Plug in the values to find the potential energy:\[PE = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot (1.6 \times 10^{-19}\,\text{C})^2}{1.0 \times 10^{-15}\, \text{m}} = 2.3 \times 10^{-13} \, \text{J}\]Since each nucleus must have half of this energy, the kinetic energy needed for one nucleus is \(1.15 \times 10^{-13}\, \text{J}\).

Step 2: Relate Kinetic Energy to Velocity

The kinetic energy for one nucleus is given by:\[KE = \frac{1}{2}mv^2\]where \(m\) is the mass of the nucleus and \(v\) is the velocity. We know \(KE = 1.15 \times 10^{-13}\, \text{J}\) and \(m = 3.34 \times 10^{-27}\, \text{kg}\). Solve for \(v\):\[1.15 \times 10^{-13} = \frac{1}{2} (3.34 \times 10^{-27})v^2\]Solving for \(v\), we have:\[v = \sqrt{\frac{2 \times 1.15 \times 10^{-13}}{3.34 \times 10^{-27}}} = 9.63 \times 10^6 \, \text{m/s}\]

Step 3: Calculate Magnetic Field for Circular Motion

When moving in a circle under a magnetic field, the magnetic force equals the centripetal force needed to keep the nuclei in circular motion. The magnetic force is given by:\[F_B = qvB\]And the centripetal force is:\[F_C = \frac{mv^2}{r}\]where \(r\) is the radius of the circle (1.25 m since the diameter is 2.50 m). Equating the two forces, we have:\[vB = \frac{mv^2}{r}\]Solving for \(B\), the magnetic field strength, we find:\[B = \frac{mv}{qr} = \frac{(3.34 \times 10^{-27}) \cdot (9.63 \times 10^6)}{(1.6 \times 10^{-19}) \cdot (1.25)} = 1.60 \, \text{T}\]

Step 4: Conclusion

The velocity needed for the deuterium nuclei to come close enough to fuse is approximately \(9.63 \times 10^6 \, \text{m/s}\). To maintain this motion in a circle of 2.50 m diameter, a magnetic field of \(1.60 \, \text{T}\) is required.

Key Concepts

Deuterium Nuclei

Deuterium nuclei, also known as heavy hydrogen nuclei, contain one proton and one neutron. This composition gives deuterium a slightly greater mass compared to the more common hydrogen isotope, which only contains a single proton.
Its nucleus is denoted as \(^2\text{H}\) or \(^2_1\text{D}\), reflecting its atomic mass number of two. Because deuterium is an isotope of hydrogen, it shares many chemical properties with its lighter counterpart but differs significantly in nuclear reactions.
The fusion of deuterium nuclei is a process that occurs in the core of stars, where conditions are extreme enough to overcome the natural repulsion between positive charges. This fusion results in the formation of helium, releasing vast amounts of energy in the process.

Coulomb Potential Energy

Coulomb potential energy (PE) refers to the energy stored in the electric field between two charged particles. It can be calculated using the formula: \[PE = \frac{k \cdot e^2}{r}\]where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\), \(e\) is the elementary charge \(1.6 \times 10^{-19}\,\text{C}\), and \(r\) is the separation distance between the charges.
In the case of nuclear fusion, this potential energy must be overcome for two like-charged nuclei to come close enough to initiate a fusion reaction. In this scenario with deuterium nuclei, the necessary separation distance is about \(1.0 \times 10^{-15}\, m\), which means the PE at this distance needs to be matched by the kinetic energy of the nuclei for fusion to occur.

Magnetic Confinement

Magnetic confinement is a technique used to control the movement of charged particles, like deuterium nuclei, within a controlled area. This is essential in nuclear fusion experiments since the particles move too fast to be contained by physical walls.
The movement of charged particles in a magnetic field follows a circular path due to the Lorentz force, which acts perpendicular to the particle's velocity.
By adjusting the strength and orientation of the magnetic field, it is possible to control their trajectories. In practice, this is achieved within devices like tokamaks, which create a toroidal (doughnut-shaped) magnetic field to confine hot plasma efficiently. In our example exercise, a magnetic field of 1.60 T is necessary to keep deuterium nuclei traveling in a 2.50-meter circular diameter path.

Strong Nuclear Force

The strong nuclear force is a fundamental force in nature, which is responsible for holding the nucleus of an atom together. It is the strongest of the four fundamental forces but only acts over extremely short distances, approximately \(10^{-15}\, m\).
This force comes into play during nuclear fusion, counteracting the electrostatic repulsion between like-charged protons within atomic nuclei.
In the fusion of deuterium nuclei, the strong nuclear force needs to pull the two nuclei together once they are close enough (beyond the reach of Coulomb potential) to allow the atomic particles to overcome their mutual repulsive force and trigger fusion. This close approach and resultant fusion are what release the vast energy observed in fusion reactions, making it a potential energy source for future clean energy technologies.