two different radioactive isotopes decay to 1...
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
Answer
# 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. Given $N = 0.1N_0$, 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. Using the formula $T_{1/2,A}=\frac{-33\ln(2)}{\ln(0.1)}$. Since $\ln(2)\approx0.693$ and $\ln(0.1)= - 2.303$, $T_{1/2,A}=\frac{-33\times0.693}{-2.303}=\frac{22.869}{2.303}\approx9.93$.
## Step3: Calculate half - life of isotope B
For isotope B, $t = 43$ days. Using the formula $T_{1/2,B}=\frac{-43\ln(2)}{\ln(0.1)}$. Substituting $\ln(2)\approx0.693$ and $\ln(0.1)= - 2.303$, $T_{1/2,B}=\frac{-43\times0.693}{-2.303}=\frac{29.799}{2.303}\approx12.94$.
## Step4: Find the difference in half - lives
$\Delta T_{1/2}=T_{1/2,B}-T_{1/2,A}\approx12.94 - 9.93\approx3$ days.
# Answer:
3 days