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
10 days
13 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 gives $\ln(0.1)=\frac{t}{T_{1/2}}\ln(\frac{1}{2})$. Then $T_{1/2}=\frac{t\ln(2)}{\ln(10)}$.

Step2: Calculate half - life of isotope A

For isotope A, $t = 33$ days. Using $T_{1/2}=\frac{t\ln(2)}{\ln(10)}$, we substitute $t = 33$: $T_{1/2A}=\frac{33\ln(2)}{\ln(10)}\approx\frac{33\times0.693}{2.303}\approx9.97\approx10$ days.

Step3: Calculate half - life of isotope B

For isotope B, $t = 43$ days. Using $T_{1/2}=\frac{t\ln(2)}{\ln(10)}$, we substitute $t = 43$: $T_{1/2B}=\frac{43\ln(2)}{\ln(10)}\approx\frac{43\times0.693}{2.303}\approx12.97\approx13$ days.

Step4: Find the difference in half - lives

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

Answer:

3 days