Question

The radioactive isotope ²⁴⁷Bk decays by a series of an alpha particle and b-particle productions, taking ²⁴⁷Bk through many transformations to end up as ²⁰⁷Pb. In the complete decay series, how many α particles and β particles are produced?

Short answer

10 alpha particles and 5 beta particles are produced.

Step-by-step solution

Determining the production of alpha particles

The overall reaction becomes:

${}_{97}{}^{247}\mathrm{Bk}\to {}_{82}{}^{207}\mathrm{Pb}+?{}_2{}^4\mathrm{He}+?{}_{-1}{}^0\mathrm{e}$

There is a change of 40 in atomic mass and a change of 15 in atomic number. In alpha decays, 10 $\left({}_2{}^4\mathrm{He}\right)$are used to decay the atom. So, the number of alpha particles produced is 10.

Determining the production of a beta particle

Bk loses 20 units in atomic number in the alpha decay. So, to compensate for the 5 units for atomic number 5, free electrons $\left({}_{-1}{}^0\mathrm{e}\right)$are added back to the atom. So, the number of beta particles produced is 5.