Question

a sample of h decays, leaving 3.1% of the original amount. how long would this take?
25
61.5
5
6.25
20
10
2
12.5
3.1
50
80
100
3
40

Explanation:

1. Assume the radioactive - decay formula is $N = N_0e^{-\lambda t}$, where $N$ is the amount of the substance at time $t$, $N_0$ is the initial amount, $\lambda$ is the decay constant.

• We know that $\frac{N}{N_0}=0.031$ (since $3.1\%=\frac{3.1}{100} = 0.031$). So, $0.031=e^{-\lambda t}$.
• Take the natural - logarithm of both sides: $\ln(0.031)=-\lambda t$.
• If we assume this is a problem with a known half - life $T_{1/2}$, and the relationship between the decay constant $\lambda$ and the half - life is $\lambda=\frac{\ln(2)}{T_{1/2}}$. For simplicity, if we assume the half - life $T_{1/2}$ is some unit value (say $T_{1/2} = 10$ for the sake of demonstration, and in a real - world scenario, we would need to know the specific isotope of hydrogen).
• First, $\ln(0.031)\approx - 3.47$.
• Since $\lambda=\frac{\ln(2)}{T_{1/2}}$ and $T_{1/2} = 10$, then $\lambda=\frac{\ln(2)}{10}\approx\frac{0.693}{10}=0.0693$.
• From $\ln(0.031)=-\lambda t$, we can solve for $t$: $t=\frac{-\ln(0.031)}{\lambda}$.
• Substitute $\ln(0.031)\approx - 3.47$ and $\lambda\approx0.0693$ into the formula for $t$: $t=\frac{3.47}{0.0693}\approx50$.

Answer:

The answer is 50.