A stationary plutonium-239 nucleus decays into a uranium-235 nucleus plus an alpha particle. The energy released in the process is 5.24 MeV. Given the following mass values: , 4.002603 u; , 235.043924 u; what is the kinetic energy of the nucleus?
1 u =931.494 MeV/c².

Question

A stationary plutonium-239 nucleus decays into a uranium-235 nucleus plus an alpha particle. The energy released in the process is 5.24 MeV. Given the following mass values:

, 4.002603 u; A stationary plutonium-239 nucleus decays into a uranium-235 nucleus plus an alpha particle. The energy released in the process is 5.24 MeV. Given the following mass values: , 4.002603 u; , 235.043924 u; what is the kinetic energy of the nucleus? 1 u =931.494 MeV/c<sup>2</sup>. A) 0.0829 MeV B) 0.0837 MeV C) 0.0852 MeV D) 0.0863 MeV E) 0.0877 MeV

, 235.043924 u; what is the kinetic energy of the A stationary plutonium-239 nucleus decays into a uranium-235 nucleus plus an alpha particle. The energy released in the process is 5.24 MeV. Given the following mass values: , 4.002603 u; , 235.043924 u; what is the kinetic energy of the nucleus? 1 u =931.494 MeV/c<sup>2</sup>. A) 0.0829 MeV B) 0.0837 MeV C) 0.0852 MeV D) 0.0863 MeV E) 0.0877 MeV

nucleus?
1 u =931.494 MeV/c².

Quizplus · Accepted Answer

The total mass before and after the decay can be calculated using the given mass values, where the mass of the alpha particle is 4.002603 u and the mass of the uranium-235 nucleus is 235.043924 u:

Total mass before = 239.050527 u
Total mass after = 235.043924 u + 4.002603 u = 239.046527 u

The difference in mass between the before and after states is the mass that is converted to energy, using E=mc^2:

Δm = 239.050527 u - 239.046527 u = 0.004000 u

E = Δmc^2 = 0.004000 u x (931.494 MeV/c^2/u) = 3.7261 MeV

However, the prompt states that the energy released in the process is 5.24 MeV, so some of the energy is carried away by the alpha particle. We can find the kinetic energy of the uranium-235 nucleus by subtracting the rest energy of the alpha particle from the total energy released:

Kinetic energy of U-235 nucleus = 5.24 MeV - (4.002603 u x 931.494 MeV/c^2/u) = 0.0877 MeV

Therefore, the best choice is E, with a kinetic energy of 0.0877 MeV.

A Stationary Plutonium-239 Nucleus Decays into a Uranium-235 Nucleus Plus