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

Answer:

A. 3 days

Explanation:

Step1: Recall decay formula

The decay formula is $N = N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$, 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, $0.1N_0=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$, which simplifies to $0.1=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$. Taking the natural logarithm of both sides: $\ln(0.1)=\frac{t}{T_{1/2}}\ln\left(\frac{1}{2}\right)$. Then $T_{1/2}=\frac{t\ln(2)}{\ln(10)}$.

Step2: Calculate half - life of isotope A

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

Step3: Calculate half - life of isotope B

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

Step4: Find the difference

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