Question

The radioactive isotope 242Cm decays by a series of a-particle and B-particle productions, taking 242Cm through many trans- formations to end up as 206Pb. In the complete decay series, how many a and ß particles are produced?

Answer 1

* There will be 4.pa decays to bring decrease in atomic number to 14 206 pb 242 i 242 Cm 96 cm - 825 This reaction undergoes 9 & - decays and 4 pe?-decays that is 9 alpha-decay and 4 beta - decay resulting in 9 alpha emissions and 4 beta emissions

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242 . Cm 8 Pb 96 Atomic number of cm (curium) is 96 Atomic number of Pb (Lead) is 82 Decrease in mass number = 242 - 206 = 36 Decrease in atomic number z 96-82 - 14 Alpha decay- Nucleus emits alpha particle ( He) ax - "Y + He 2-2 In an alpha decay atomic number decreases by 2 & mass number decreases by Beta decay (B) - Nucleus emits an electron (e-) and an electron antineutrino (Ve) - X - Y + en + ve DI DI In, a B decay atomic number increases by 1 & mass number remains unchanged ::. In this question, the change is mass number is solely due to a-decay => Number of a-decay = 36 = 9 There are 9 - decays => Change ( decrease) is atomic number is all a-decays = 9x2 - 18 .. But decrease in atomic number is only 14