Problem 102 - University Physics with Modern Physics

Question

\(\mathrm{A}^{232} \mathrm{Th}\) (thorium) nucleus at rest decays to a \(^{228} \mathrm{Ra}\) (radium) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is \(6.54 \times 10^{-13} \mathrm{J}\) . An alpha particle has 1.76\(\%\) of the mass of a \(^{228} \mathrm{Ra}\) nucleus. Calculate the kinetic energy of (a) the recoiling 228 nucleus and (b) the alpha particle.

Step-by-Step Solution

Answer

The kinetic energy of the recoiling nucleus is \(1.13 \times 10^{-14} \text{ J}\), and of the alpha particle is \(6.43 \times 10^{-13} \text{ J}\).

Step 1: Understanding Conservation of Momentum

The initial momentum of the thorium nucleus is zero since it's at rest. After decay, the momentum of the radium nucleus and the alpha particle must sum to zero to conserve momentum.If the kinetic energy of the radium nucleus is denoted as \(K_{Ra}\) and the kinetic energy of the alpha particle is \(K_{\alpha}\), the momentum of each can be related to their masses \(M_{Ra}\) and \(M_{\alpha}\) respectively:\[M_{Ra} v_{Ra} = M_{\alpha} v_{\alpha}\]Since kinetic energy \(K\) is \(\frac{1}{2} mv^2\), this relationship can be used to express the kinetic energies in terms of mass ratios.

Step 2: Establishing Mass Ratio

From the problem statement, we know the mass of the alpha particle is 1.76\(\%\) of the radium nucleus. Let’s denote the mass of the radium nucleus \(M_{Ra}\) and the mass of the alpha particle \(M_{\alpha}\) as follows:\[M_{\alpha} = 0.0176 \times M_{Ra}\]

Step 3: Using Kinetic Energy Relation

The total kinetic energy from decay is the sum of the kinetic energy of the radium nucleus and the alpha particle:\[K_{Ra} + K_{\alpha} = 6.54 \times 10^{-13} \text{ J} \]From conservation of momentum and substituting in terms of kinetic energy, we have:\[K_{Ra} = \frac{M_{\alpha}}{M_{Ra}} K_{\alpha}\]Substituting \(M_{\alpha} = 0.0176 \times M_{Ra}\), we get:\[K_{Ra} = 0.0176 K_{\alpha}\]

Step 4: Solving for Individual Kinetic Energies

Replacing \(K_{Ra}\) in the total energy equation:\[0.0176 K_{\alpha} + K_{\alpha} = 6.54 \times 10^{-13}\]\[1.0176 K_{\alpha} = 6.54 \times 10^{-13}\]\[K_{\alpha} = \frac{6.54 \times 10^{-13}}{1.0176}\]\[K_{\alpha} \approx 6.43 \times 10^{-13} \text{ J}\]Using \(K_{Ra} = 0.0176 K_{\alpha}\):\[K_{Ra} = 0.0176 \times 6.43 \times 10^{-13}\]\[K_{Ra} \approx 1.13 \times 10^{-14} \text{ J}\]

Step 5: Final Result

We have determined that the kinetic energy of the recoiling radium nucleus (\(^{228}Ra\)) is approximately \(1.13 \times 10^{-14} \text{ J}\), and the kinetic energy of the alpha particle is approximately \(6.43 \times 10^{-13} \text{ J}\).

Key Concepts

Conservation of Momentum

In the world of physics, the conservation of momentum is a fundamental principle. Imagine our scenario: a thorium nucleus initially at rest. When it decays into a radium nucleus and an alpha particle, that initial condition tells us something important. Since the thorium nucleus started with no movement, its momentum was zero.
So, what does this mean when the decay occurs? The resultant particles, in this case, the newly formed radium nucleus and the alpha particle, must have momentums that cancel each other out. This ensures that the total momentum post-decay remains zero, exactly as it was before the nuclear decay began.
When applying this principle, the equation used is:

• \( M_{Ra} v_{Ra} = M_{\alpha} v_{\alpha} \)

This simple relationship tells us that the product of mass and velocity (akin to momentum) for each particle must be equal. Thus, the sum of their momentums remains at a nice, neat zero.

Kinetic Energy

Kinetic energy is all about the energy of motion. During decay, the thorium nucleus bravely splits into two chunks: a radium nucleus and an alpha particle. Each takes a share of the available energy, moving apart rapidly.
What's important here? The total kinetic energy from the decay is already known—in this case, physics gives us a value of \(6.54 \times 10^{-13} \text{ J}\). This is the sum of the separate energies of the radium nucleus and the alpha particle.
Using conservation laws, we express this as:

• \( K_{Ra} + K_{\alpha} = 6.54 \times 10^{-13} \text{ J} \)

Where:

• \( K_{Ra} \) is the kinetic energy of the radium nucleus.
• \( K_{\alpha} \) is the kinetic energy of the alpha particle.

From the momentum and the known mass ratio, you can calculate the individual energy values, for both the massive radium nucleus and the lighter alpha particle, maintaining the overall energy account balanced.

Alpha Particle Emission

Alpha particle emission is a common mode of nuclear decay for heavy, unstable nuclei like thorium. An alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons. It’s pretty massive for a tiny particle, but quite light compared to the big player like the radium nucleus.
In our specific example, the alpha particle's mass is only 1.76% of that of the radium nucleus. This proportion is key when figuring out how energy from decay splits between the radium nucleus and the alpha particle.
With its relatively tiny mass, the alpha particle gets a more significant portion of the kinetic energy. Why? Simply put, it's just like how a small rock will fly further than a heavy boulder if launched with the same force.
In numerical terms, using the given mass ratio: \( M_{\alpha} = 0.0176 \times M_{Ra} \), and the relationships involving energy and momentum, we're able to accurately determine how much energy winds up with the agile little alpha particle after the decay process is complete.