Question

If the half-life of an isotope is 3 years, what fraction of the isotope remains after 4 years? (put fraction in as decimal)

Answer 1

Solution. Using radioactive decay law

$$N(t)=N_0\times e^{-\lambda t}=N_0\times e^{-\frac{tln 2}{T}}=N_0\times2^{-\frac{t}{T}}$$

where N is the number of non-decaying isotopes; N₀ is initial number of isotopes; t=4 years is time; T=3 years is half life.

Therefore, the fraction of the isotope that remains after 4 years

$$\frac {N(4)}{N_0}=2^{-\frac{4}{3}} \approx 0.4$$

Answer. 0.4.