two different radioactive isotopes decay to 10% of their respective original amounts. isotope a does this in…

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 $N = 0.1N_0$. So, $0.1N_0=N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, 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 $T_{1/2}=\frac{t\ln(2)}{\ln(10)}$.

Step2: Calculate half - life of isotope A

For isotope A, $t = 33$ days. Using the formula $T_{1/2,A}=\frac{33\ln(2)}{\ln(10)}$. Since $\ln(2)\approx0.693$ and $\ln(10)\approx2.303$, $T_{1/2,A}=\frac{33\times0.693}{2.303}=\frac{22.869}{2.303}\approx9.93$.

Step3: Calculate half - life of isotope B

For isotope B, $t = 43$ days. Using the formula $T_{1/2,B}=\frac{43\ln(2)}{\ln(10)}$. Substituting $\ln(2)\approx0.693$ and $\ln(10)\approx2.303$, $T_{1/2,B}=\frac{43\times0.693}{2.303}=\frac{29.799}{2.303}\approx12.94$.

Step4: Calculate the difference

The difference $\Delta T_{1/2}=T_{1/2,B}-T_{1/2,A}\approx12.94 - 9.93\approx3$ days.

Answer:

3 days