The kinetic energy (in keV) of the alpha particle, when the nucleus 210 84 Po at rest undergoes alpha decay, is

To find the kinetic energy of the alpha particle emitted during the alpha decay of the polonium-210 (210Po) nucleus, we need to understand a few key concepts about nuclear decay and energy conservation. In alpha decay, a heavy nucleus loses an alpha particle, which consists of 2 protons and 2 neutrons, essentially a helium nucleus.

Understanding the Decay Process

The equation for the alpha decay of polonium-210 can be written as follows:

210₈₄Po → 206₈₂Pb + 4₂He

In this reaction, the polonium-210 nucleus transforms into lead-206 (206Pb) and emits an alpha particle (4He). The conservation of momentum and energy principles are crucial here.

Applying Conservation of Energy

Before the decay, the polonium nucleus is at rest, meaning its initial kinetic energy is zero. After the decay, the total kinetic energy of the system must also account for the kinetic energy of the lead nucleus and the emitted alpha particle. However, since the lead nucleus will recoil, we focus on the alpha particle's kinetic energy.

Energy Release in Decay

The energy released during alpha decay can be calculated using the mass difference between the reactants and products. The energy release is given by:

ΔE = (mass_initial - mass_final) * c²

Where ΔE is the energy released, mass_initial is the mass of the original polonium nucleus, and mass_final is the sum of the masses of the lead nucleus and the alpha particle. The speed of light (c) in this equation is approximately 3 x 10⁸ m/s.

Mass Values

• Mass of 210Po ≈ 209.9828 u
• Mass of 206Pb ≈ 205.9745 u
• Mass of 4He ≈ 4.0026 u

Calculating the Mass Defect

The mass defect can be calculated as follows:

mass_defect = mass_initial - (mass_final for Pb + mass_final for He)

mass_defect = 209.9828 u - (205.9745 u + 4.0026 u) = 209.9828 u - 209.9771 u = 0.0057 u

Converting Mass Defect to Energy

Now, we convert this mass defect into energy using Einstein's equation, where 1 atomic mass unit (u) is equivalent to about 931.5 MeV:

ΔE = 0.0057 u * 931.5 MeV/u ≈ 5.30 MeV

Kinetic Energy of the Alpha Particle

In alpha decay, the majority of the energy released goes to the emitted alpha particle due to its relatively small mass compared to the recoiling lead nucleus. Therefore, the kinetic energy of the alpha particle is approximately equal to the energy released during the decay:

Kinetic energy of the alpha particle ≈ 5.30 MeV

Final Result in keV

Since 1 MeV is equivalent to 1000 keV, we can convert the energy:

5.30 MeV = 5300 keV

In summary, the kinetic energy of the alpha particle emitted during the alpha decay of polonium-210 is approximately 5300 keV.