Question

an 80 - gram sample of h decays, leaving 2.5 g of h. how long would this take? 1 12.3 100 3 10 20 2 5 80 6.25 40 50 61.5 12.5

Explanation:

Step1: Assume radioactive - decay formula

The radioactive - 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. Let's assume the half - life $T_{1/2}=1$ for simplicity (if the half - life is not given, we can find the number of half - lives passed). Here, $N_0 = 80$ g and $N = 2.5$ g.

Step2: Substitute values into the formula

We have $2.5=80(\frac{1}{2})^{\frac{t}{T_{1/2}}}$. First, divide both sides by 80: $\frac{2.5}{80}=(\frac{1}{2})^{\frac{t}{T_{1/2}}}$. Since $\frac{2.5}{80}=\frac{25}{800}=\frac{1}{32}$, the equation becomes $\frac{1}{32}=(\frac{1}{2})^{\frac{t}{T_{1/2}}}$.

Step3: Rewrite the fraction

We know that $\frac{1}{32}=\frac{1}{2^5}=(\frac{1}{2})^5$. So, $(\frac{1}{2})^5 = (\frac{1}{2})^{\frac{t}{T_{1/2}}}$. If $T_{1/2}=1$, then $t = 5$.

Answer:

5