The Binding Energy and Energy Release of Polonium-210

What is the binding energy of Polonium-210 and how much energy is released during its alpha decay?

a) Calculate the binding energy of polonium-210.

b) Calculate the energy released during the alpha decay of polonium-210.

Answer:

a) The binding energy of Polonium-210 is 91.25 MeV.

b) The energy released during the alpha decay of Polonium-210 is 5.86 MeV.

Polonium-210 has a binding energy of 91.25 MeV and releases 5.86 MeV of energy during its alpha decay.

Binding Energy of Polonium-210:

Binding energy is the energy required to break a nucleus into its individual protons and neutrons. Polonium-210 has an atomic mass of 209.9828 u, with 84 protons and an atomic number of 84.

The number of neutrons in a polonium atom is given by: 209.9828 u - (84 protons × 1.00728 u/proton) = 126.9255 u.

The mass defect of Polonium-210 is calculated as: (126.9255 u × 1.00867 u) + (84 protons × 1.00728 u/proton) - 209.9828 u = 0.0983 u.

Using Einstein's equation, E=mc², the binding energy is calculated as: BE = (0.0983 u) × (931.5 MeV/u) = 91.25 MeV.

Energy Released During Alpha Decay:

During alpha decay, Polonium-210 decays into lead-206 by emitting an alpha particle, which is a helium nucleus. The mass defect for alpha decay is calculated as: 209.9828 u - (205.974 u + 4.0015 u) = 0.0063 u.

The energy released during alpha decay is determined by: Energy = (0.0063 u) × (931.5 MeV/u) = 5.86 MeV.

Alpha decay is a type of radioactive decay where an atomic nucleus emits an alpha particle, releasing energy in the process. The binding energy of Polonium-210 and the energy released during its alpha decay are crucial factors in nuclear physics.