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.

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.

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.

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

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.