16 g of a radio active substance is reduced to 0.5 g after 1 hour. The half life of the radioactive substance in minutes is

Answer

To find the half-life of a radioactive substance given that 16 g reduces to 0.5 g in 1 hour, we can follow these steps:

### Step 1: Identify Initial and Final Amounts
- Initial amount (A₀) = 16 g
- Final amount (A) = 0.5 g
- Time (t) = 1 hour = 60 minutes

### Step 2: Use the First-Order Reaction Formula
For a first-order reaction, the rate constant (k) is given by the formula:
\[ k = \frac{2.303}{T} \log \left( \frac{A_0}{A} \right) \]

### Step 3: Substitute the Values into the Formula
Substituting the known values into the formula:
\[ k = \frac{2.303}{60} \log \left( \frac{16}{0.5} \right) \]

### Step 4: Calculate the Logarithm
Calculate the fraction:
\[ \frac{16}{0.5} = 32 \]
Now, find the logarithm:
\[ \log(32) = \log(2^5) = 5 \log(2) \]
Using the value of \(\log(2) \approx 0.3010\):
\[ \log(32) = 5 \times 0.3010 = 1.505 \]

### Step 5: Substitute Back to Find k
Now substitute this back into the equation for k:
\[ k = \frac{2.303}{60} \times 1.505 \]

### Step 6: Calculate k
Calculating k:
\[ k \approx \frac{2.303 \times 1.505}{60} \]
\[ k \approx \frac{3.469515}{60} \approx 0.05782525 \]

### Step 7: Calculate Half-Life
The half-life (t₁/₂) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of k:
\[ t_{1/2} = \frac{0.693}{0.05782525} \]

### Step 8: Final Calculation
Calculating the half-life:
\[ t_{1/2} \approx 12 minutes \]

### Conclusion
The half-life of the radioactive substance is approximately **12 minutes**.

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