Question

When a bismuth-208 nucleus at rest decays, thallium-204 is produced, along with an alpha particle (helium- 4 nucleus). The mass numbers of bismuth-208, thallium- 204 , and helium- 4 are 208,204 , and 4 , respectively. (The mass number represents the total number of protons and neutrons in the nucleus.) The kinetic energy of the thallium nucleus is a) equal to that of the alpha particle. b) less than that of the alpha particle. c) greater than that of the alpha particle.

Short answer

Question: In a decay process, a bismuth-208 nucleus at rest decays into a thallium-204 nucleus and an alpha particle (helium-4 nucleus). The kinetic energy of the thallium-204 nucleus is: a) equal to that of the alpha particle b) less than that of the alpha particle c) greater than that of the alpha particle Answer: b) less than that of the alpha particle.

Step-by-step solution

Understand the decay process

During the decay process, a bismuth-208 nucleus at rest decays into a thallium-204 nucleus and an alpha particle, also known as a helium-4 nucleus.

Apply conservation of momentum principle

In this decay process, linear momentum is conserved. Since the initial linear momentum of the bismuth-208 nucleus is zero (it's at rest), the total linear momentum of the thallium-204 nucleus and the alpha particle must also be zero. Mathematically, this can be written as: m_T * v_T = m_α * v_α where m_T and v_T are the mass and velocity of the thallium-204 nucleus, and m_α and v_α are the mass and velocity of the alpha particle, respectively.

Relate kinetic energy to momentum

The kinetic energy of a particle is given by the formula: K = (1/2) * m * v^2 We want to compare the kinetic energies of the thallium-204 nucleus and the alpha particle. Let's find an expression relating their velocities using the conservation of momentum equation from step 2: v_α = (m_T / m_α) * v_T

Compare the kinetic energies

Now that we have an expression relating the velocities of the thallium-204 nucleus and the alpha particle, we can compare their kinetic energies. The kinetic energy of the thallium-204 nucleus is: K_T = (1/2) * m_T * v_T^2 The kinetic energy of the alpha particle, using the expression for v_α that we found in step 3, is: K_α = (1/2) * m_α * ((m_T / m_α) * v_T)^2 K_α = (1/2) * m_α * (m_T^2 / m_α^2) * v_T^2 K_α = (m_T^2 / (2 * m_α)) * v_T^2 Now, we can compare the ratios of their kinetic energies: (K_T / K_α) = (m_T * v_T^2) / ((m_T^2 / (2 * m_α)) * v_T^2) After canceling out common factors, we get: (K_T / K_α) = (2 * m_α) / m_T Since m_T (mass of thallium-204) > m_α (mass of alpha particle): (K_T / K_α) < 1 Therefore, the kinetic energy of the thallium-204 nucleus is less than that of the alpha particle. The correct answer is: b) less than that of the alpha particle.

Key concepts

Conservation of Momentum

In the fascinating world of nuclear decay, conservation of momentum plays a crucial role. Imagine a bismuth-208 nucleus initially at rest. When it undergoes decay, it releases a thallium-204 nucleus and an alpha particle. In such a scenario, the principle of conservation of linear momentum is vital. It assures us that the total momentum before and after the decay remains the same.

Since the bismuth-208 nucleus is at rest initially, its momentum is zero. Hence, the combined momentum of the thallium-204 nucleus and the alpha particle should also be zero post-decay. This is mathematically expressed as:

• \( m_T \cdot v_T = m_\alpha \cdot v_\alpha \),

Here, \( m_T \) and \( v_T \) are the mass and velocity of the thallium-204 nucleus, while \( m_\alpha \) and \( v_\alpha \) represent those of the alpha particle. By conserving momentum, we can intuitively infer and calculate the possible velocities of the decay products. This understanding acts as a window into the dynamics of nuclear decay.

Kinetic Energy

Kinetic energy is the energy of motion, and during nuclear decay like the one experienced by a bismuth-208 nucleus, kinetic energy becomes particularly enlightening. After decay, as a thallium-204 nucleus and an alpha particle move away from each other, each possesses kinetic energy.

The kinetic energy of an object is given by the formula:

• \( K = \frac{1}{2} \cdot m \cdot v^2 \),

where \( m \) is the mass and \( v \) the velocity of the object. When it comes to comparing the two decay products—thallium and alpha particle—we use the relation we derived from momentum conservation to understand their velocity ratio.

Expressing the velocities in terms of each other allows us to evaluate and compare their kinetic energies effectively:

• \( K_\alpha = \left( \frac{m_T^2}{2 \cdot m_\alpha} \right) \cdot v_T^2 \).

Given the equations, and knowing that the mass of thallium (\( m_T \)) is greater than that of the alpha particle (\( m_\alpha \)), it's evident that the kinetic energy of thallium is less than that of the alpha particle. Thus, this informs us about the kinetic energy distribution in the decay process.

Alpha Particle Emission

Alpha particle emission is one form of nuclear decay where a parent nucleus ejects an alpha particle, which is essentially a helium-4 nucleus. This means the alpha particle contains 2 protons and 2 neutrons, making it quite a heavy particle compared to electrons or neutrinos.

During this emission from a bismuth-208 nucleus, the resultant products are a thallium-204 nucleus and the emitted alpha particle. The mass number of alpha particles is 4, highlighting their significance in changing the structure of the original nucleus. Their emission is a key factor in reducing the mass number and atomic number of the decaying nucleus.

Understanding alpha particles is integral to grasping nuclear processes:

• The emission reduces the mass number of the parent nucleus by 4.
• The atomic number decreases by 2.

This type of decay reflects a fundamental nuclear transformation, making the study of alpha particles essential to nuclear physics. It helps explain nuclear stability and the natural occurrence of radioactive elements, enriching the bigger picture of atomic physics.