To find the half-life of a radioactive substance, we use the formula for exponential decay:

$N(t) = N_0 e^{-kt}$

Where:

• $N(t)$ is the amount of substance at time $t$,
• $N_0$ is the initial amount,
• $k$ is the decay constant,
• $t$ is time.

We know that the substance is reduced by 22% in 5 hours. So, 78% (100% - 22%) of the substance remains after 5 hours:

$\frac{N(t)}{N_0} = 0.78$

At $t = 5$ hours, the equation becomes:

$0.78 = e^{-k \cdot 5}$

Taking the natural logarithm of both sides:

$\ln(0.78) = -5k$

Solving for $k$:

$k = \frac{-\ln(0.78)}{5}$

Now, the half-life $T_{\frac{1}{2}}$ is related to the decay constant $k$ by the formula:

$T_{\frac{1}{2}} = \frac{\ln(2)}{k}$

Let's calculate this step by step.The half-life of the radioactive substance is approximately 13.95 hours.

Would you like further details on this calculation or have any questions?

Here are five related questions to deepen your understanding:

1. What is the relationship between half-life and decay constant?
2. How does the half-life change if the substance decays faster?
3. Can this method be used for substances with different decay percentages?
4. How would the formula change if the substance had grown instead of decaying?
5. What happens to the remaining quantity as time approaches infinity?

Tip: Understanding the logarithmic function is crucial for solving decay-related problems, as it helps in determining the time and rate of decay accurately.