The U-238 nucleus decays to form Pb-206 by α and β decays.
-Calculate the number of β decays.

Question

Quizplus · Accepted Answer

6In the decay of U-238 to Pb-206, the process involves both alpha (α) and beta (β) decays. Each alpha decay decreases the atomic number by 2 and the mass number by 4. To go from uranium-238 to lead-206, a series of these decays happens, reducing the mass number by a total of 32 (238 - 206 = 32). Since each alpha decay reduces the mass number by 4, there are 8 alpha decays in total (32/4 = 8).Alpha decay reduces the atomic number by 2, so 8 alpha decays reduce it by 16 (2*8 = 16). The atomic number of uranium (U) is 92, and the atomic number of lead (Pb) is 82. The difference in atomic numbers is 10 (92 - 82 = 10), but the alpha decays account for a reduction of 16. This discrepancy is because alpha decays only account for changes in the atomic and mass numbers, not the conversion of neutrons to protons, which is achieved through beta decay.Beta decay increases the atomic number by 1 without changing the mass number. To account for the atomic number of lead being 10 less than uranium while only reducing it by 16 through alpha decays, 6 beta decays are needed to increase the atomic number back up by 6 (82 + 6 = 88, but since we're counting the reduction from 92, it's effectively a net change of 10 with the alpha decays considered). Therefore, there are 6 beta decays involved in the process of U-238 decaying to Pb-206.

The U-238 Nucleus Decays to Form Pb-206 by α and β