| |||
|
Alpha decay of Rn-222 to Po-218 radionuclide can be represented by the following equation:
86Rn222 = 84Po218 + 2He4 + Q The emitted kinetic energy of alpha particle (2He4) is equal to 5.4897 MeV. Alpha decay energy of 86Rn222 is: E(alpha) = 5.4897 MeV + (5.4897 MeV x 4 amu)/218 amu = 5.5904 MeV amu = atomic mass unit In addition to an alpha particle, in this decay a gamma ray with 0.51 MeV of energy is emitted, which indicates that alpha particle is emitted with a low kinetic energy, leaving to 84Po218 in an excited state. |
| GARZON HOME |
