Question

When 234/91 Pa decays, it emits five alpha particles and two beta particles in seven steps. What is the final product of this series of decay steps?

Answer 1

An alpha particle has mass number 4 and atomic number 2. Thus, loss of an alpha particle results in decrease in mass number by 4 units and decrease in atomic number by 2 units.

Therefore, loss of 5 alpha particles will result in decrease in mass number by 5X4 = 20 units and decrease in atomic number by 5X2 =10 units.

Therefore, after loss of five alpha particles from 234/91 Pa, new element will have mass number 234-20 = 214 and atomic number 91 – 10 = 81.

A beta particle loss results in increase in atomic number by one unit and no change in mass number. Thus, loss of 2 beta particles results in increase in atomic number by 2 X 2 = 4 units.

Therefore, after loss of 2 beta particles the change in atomic number will be 81 +2 = 83.

Hence, final product of seven decay steps will be having atomic number of 83 and mass number of 214. Since, Bismuth (Bi) has atomic number 83, therefore, final product is isotope of bismuth, 214/83 Bi