Question

two different radioactive isotopes decay to 10% of their respective original amounts. isotope a does this in 33 days, while isotope b does this in 43 days. what is the approximate difference in the half - lives of the isotopes? 3 days 13 days 10 days 33 days

Explanation:

Step1: Recall decay formula

The decay formula is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N$ is the final amount, $N_0$ is the initial amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. When $N = 0.1N_0$, we have $0.1N_0=N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, which simplifies to $0.1 = (\frac{1}{2})^{\frac{t}{T_{1/2}}}$.
Taking the natural logarithm of both sides: $\ln(0.1)=\frac{t}{T_{1/2}}\ln(\frac{1}{2})$. Then $T_{1/2}=\frac{-t\ln(2)}{\ln(0.1)}$.

Step2: Calculate half - life of isotope A

For isotope A, $t = 33$ days. $T_{1/2A}=\frac{-33\ln(2)}{\ln(0.1)}\approx\frac{-33\times0.693}{- 2.303}\approx9.97\approx10$ days.

Step3: Calculate half - life of isotope B

For isotope B, $t = 43$ days. $T_{1/2B}=\frac{-43\ln(2)}{\ln(0.1)}\approx\frac{-43\times0.693}{-2.303}\approx12.97\approx13$ days.

Step4: Find the difference

The difference $\Delta T_{1/2}=T_{1/2B}-T_{1/2A}\approx13 - 10=3$ days.

Answer:

3 days