Question

A radioactive substance has a half-life of 92 years. How long will it take for 65% of the substance to decay?

Answer 1

To determine the time it takes for 65% of a radioactive substance to decay, we can use the formula for exponential decay, which is given by:

\[
N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}
\]

Where:
- \(N(t)\) is the amount of substance remaining at time \(t\),
- \(N_0\) is the initial amount of substance,
- \(T_{1/2}\) is the half-life, and
- \(t\) is the time that has elapsed.

If 65% of the substance has decayed, that means 35% remains. Thus, we have:

\[
N(t) = 0.35 N_0
\]

Substituting this into the decay formula gives:

\[
0.35 N_0 = N_0 \left( \frac{1}{2} \right)^{\frac{t}{92}}
\]

Dividing both sides by \(N_0\) (assuming \(N_0 \neq 0\)), we get:

\[
0.35 = \left( \frac{1}{2} \right)^{\frac{t}{92}}
\]

Next, we take the logarithm of both sides:

\[
\log(0.35) = \log\left( \left( \frac{1}{2} \right)^{\frac{t}{92}} \right)
\]

Using the power rule of logarithms:

\[
\log(0.35) = \frac{t}{92} \log\left( \frac{1}{2} \right)
\]

Solving for \(t\):

\[
t = 92 \cdot \frac{\log(0.35)}{\log\left(\frac{1}{2}\right)}
\]

Calculating the logarithms, we have:

\[
\log(0.35) \approx -0.455
\]
\[
\log\left(\frac{1}{2}\right) \approx -0.301
\]

Substituting these values into the equation:

\[
t \approx 92 \cdot \frac{-0.455}{-0.301}
\]
\[
t \approx 92 \cdot 1.51 \approx 139.32 \text{ years}
\]

Thus, it will take approximately **139.32 years** for 65% of the radioactive substance to decay.