A radioactive decay series that begins with 232 90 T h ends with formation of th

Question

A radioactive decay series that begins with 23290Th ends with formation of the stable nuclide 20882Pb. Part A How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays? A radioactive decay series that begins with 23290Th ends with formation of the stable nuclide 20882Pb. A radioactive decay series that begins with 23290Th ends with formation of the stable nuclide 20882Pb. A radioactive decay series that begins with 23290Th ends with formation of the stable nuclide 20882Pb. Part A How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

Explanation / Answer

An alpha decay changes the atomic number by -2 and the atomic mass number by -4 (an alpha particle is a He-4 nucleus containing 2 protons and 2 neutrons).

A beta-minus decay changes the atomic number by +1, and does not change the atomic mass number (a beta-minus particle is an electron).

From the above, it should be clear that only alpha decays change the atomic mass number, so in going from Th-232 to Pb-208, the mass number changes by 232-208 = 24 units. Each alpha decay reduces the mass by 4 units, so there must be 6 alpha decays in this chain.

Each of those alpha decays will change the atomic number by -2, so in the absence of any beta decays, we would end up with a nucleus with atomic number 90 - 2*6 = 78. In fact, Pb has an atomic number of 82, which is 4 units higher than 78. Each beta-minus decay increases the atomic number by 1, so there must be 4 beta decays in this chain.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.