The half-life of a radioactive substance is 40 days. (i)The time taken for 3
5
of the original

number of atoms to disintegrate and (ii) The time for 3
4
of the original number of atoms to

remain unchanged.

To solve the problem, we'll use the concept of half-life and the exponential decay formula. The number of remaining atoms \( N(t) \) at time \( t \) can be given by the formula:

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

where:
- \( N_0 \) is the initial number of atoms,
- \( T_{1/2} \) is the half-life (40 days in this case),
- \( t \) is the time elapsed.

### (i) Time taken for \(\frac{3}{5}\) of the original number of atoms to disintegrate

First, we need to find the time when \( \frac{3}{5} \) of the original atoms have disintegrated. This means that \( \frac{2}{5} \) of the original atoms remain. Therefore, we set \( N(t) = \frac{2}{5} N_0 \) and solve for \( t \).

\[
\frac{2}{5} N_0 = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{40}}
\]

\[
\frac{2}{5} = \left(\frac{1}{2}\right)^{\frac{t}{40}}
\]

Take the natural logarithm on both sides:

\[
\ln\left(\frac{2}{5}\right) = \frac{t}{40} \ln\left(\frac{1}{2}\right)
\]

\[
t = 40 \times \frac{\ln\left(\frac{2}{5}\right)}{\ln\left(\frac{1}{2}\right)}
\]

Calculating this:

\[
t = 40 \times \frac{\ln(0.4)}{\ln(0.5)} \approx 40 \times \frac{-0.9163}{-0.6931} \approx 52.88 \text{ days}
\]

### (ii) Time for \(\frac{3}{4}\) of the original number of atoms to remain unchanged

Here, we want the time \( t \) when \( \frac{3}{4} \) of the original number of atoms remain, i.e., \( N(t) = \frac{3}{4} N_0 \).

\[
\frac{3}{4} N_0 = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{40}}
\]

\[
\frac{3}{4} = \left(\frac{1}{2}\right)^{\frac{t}{40}}
\]

Taking the natural logarithm on both sides:

\[
\ln\left(\frac{3}{4}\right) = \frac{t}{40} \ln\left(\frac{1}{2}\right)
\]

\[
t = 40 \times \frac{\ln\left(\frac{3}{4}\right)}{\ln\left(\frac{1}{2}\right)}
\]

Calculating this:

\[
t = 40 \times \frac{\ln(0.75)}{\ln(0.5)} \approx 40 \times \frac{-0.2877}{-0.6931} \approx 16.6 \text{ days}
\]

### Summary:
1. The time taken for \(\frac{3}{5}\) of the original atoms to disintegrate is approximately **52.88 days**.
2. The time taken for \(\frac{3}{4}\) of the original atoms to remain unchanged is approximately **16.6 days**.

Would you like more details on any part of the solution? 

Here are some related questions:
1. How do you derive the exponential decay formula?
2. What is the significance of the natural logarithm in solving half-life problems?
3. How does the half-life change if a substance has a different decay constant?
4. How can you model radioactive decay using differential equations?
5. What is the difference between half-life and mean lifetime?

**Tip:** When solving decay problems, always ensure your logarithms are correctly handled—missteps can lead to significant errors in time calculations.

Learn how to solve a radioactive decay problem using the half-life concept. Find out the time taken for 3/5 of the original number of atoms to disintegrate and for 3/4 of the original number of atoms to remain unchanged. Detailed step-by-step solutions and calculations included.

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