A particular radioisotope X with a half-life of 4 years decays into the stable isotope Y. At a particular time, a sample contains 32 x 10²º atoms of nuclide X and 4 x 10²º atoms of nuclide Y. How many atoms of nuclide Y will be present in the sample 8 years later?

Answer:

3) F

Number of half-lives after 8 years = 8/4 = 2 half-lives
Number of atoms of X left after 8 years = 32×10²⁰ ÷ (2×2) = 8×10²⁰

We know all of the X atoms decay into isotope Y.
Number of X atoms that decay = 32×10²⁰ – 8×10²⁰ = 24×10²⁰
We know that initial number of Y atoms present were 4×10²⁰
Number of Y atoms present after 8 years = 24+4 = 28×10²⁰