Activity of a substance changes from `700 s^(–1)` to `500 s^(–1)` in 30 minute. Find its half-life in minutes

Options

A. 66
B. 52
C. 56
D. 62

Explanation

To find the half-life of a substance whose activity changes from \(700 \, s^{-1}\) to \(500 \, s^{-1}\) in 30 minutes, we can follow these steps:

### Step 1: Understand the relationship between activity and the number of nuclei
The activity \(A\) of a radioactive substance is directly proportional to the number of undecayed nuclei \(N\):
\[ A \propto N \]
Thus, if the activity changes, the ratio of the initial and final activities can be used to find the ratio of the initial and final numbers of nuclei.

### Step 2: Set up the equation for activity
Let:
- \(A_0 = 700 \, s^{-1}\) (initial activity)
- \(A = 500 \, s^{-1}\) (final activity)
- \(t = 30 \, \text{minutes} = 1800 \, \text{seconds}\)

Using the relationship between activity and number of nuclei, we can write:
\[ \frac{A_0}{A} = \frac{N_0}{N} \]
Substituting the values:
\[ \frac{700}{500} = \frac{N_0}{N} \]
This simplifies to:
\[ \frac{N_0}{N} = \frac{7}{5} \]

### Step 3: Use the decay formula
The decay of a radioactive substance can be described by the equation:
\[ \ln\left(\frac{N_0}{N}\right) = \lambda t \]
Substituting the ratio we found:
\[ \ln\left(\frac{7}{5}\right) = \lambda \cdot 1800 \]

### Step 4: Solve for the decay constant \(\lambda\)
Rearranging the equation gives:
\[ \lambda = \frac{\ln\left(\frac{7}{5}\right)}{1800} \]

### Step 5: Calculate the half-life
The half-life \(t_{1/2}\) is related to the decay constant by the formula:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting the expression for \(\lambda\):
\[ t_{1/2} = \frac{\ln(2)}{\frac{\ln\left(\frac{7}{5}\right)}{1800}} = \frac{1800 \cdot \ln(2)}{\ln\left(\frac{7}{5}\right)} \]

### Step 6: Compute the numerical value
Now, we can calculate the numerical value of the half-life. Using approximate values:
- \(\ln(2) \approx 0.693\)
- \(\ln\left(\frac{7}{5}\right) \approx \ln(1.4) \approx 0.336\)

Substituting these values:
\[ t_{1/2} \approx \frac{1800 \cdot 0.693}{0.336} \approx 61.8 \, \text{minutes} \]

### Conclusion
Thus, the half-life of the substance is approximately \(62\) minutes.

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