Answer :
To find the disintegration (decay) constant and the time taken for 25% of the radioactive sample's activity to disappear, follow the steps below:
### Step 1: Calculate the Decay Constant
The decay constant (\( \lambda \)) is related to the half-life (\( t_{1/2} \)) by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Given that the half-life \( t_{1/2} \) is 8.3 x 10^4 years, we can substitute this into the formula:
\[ \lambda = \frac{\ln(2)}{8.3 \times 10^4 \text{ years}} \]
### Step 2: Compute the Decay Constant
Using a calculator:
\[ \lambda \approx \frac{0.693}{8.3 \times 10^4} \approx 8.35 \times 10^{-6} \text{ per year} \]
This is the decay constant in units of years^-1.
### Step 3: Calculate the Time for 25% Activity to Disappear
To find the time \( t \) it takes for the activity to reduce to 25%, we first need to realize that if only 25% remains, then 75% has disappeared. So, what we're really looking for is the time taken for the activity to decay from 100% to 25%, or equivalently to decay by 75%.
The formula that relates the remaining fraction of a radioactive sample to the decay constant and time is given by:
\[ N(t) = N_0 e^{-\lambda t} \]
Where:
- \( N(t) \) is the remaining amount of substance after time \( t \),
- \( N_0 \) is the initial amount of substance,
- \( e \) is the base of the natural logarithm,
- \( \lambda \) is the decay constant, and
- \( t \) is time in years.
We can rearrange this to solve for \( t \) as follows:
\[ t = \frac{\ln(\frac{N(t)}{N_0})}{-\lambda} \]
Since \( N(t) \) is 25% of \( N_0 \), we have \( \frac{N(t)}{N_0} = 0.25 \). Plugging in the decay constant and this ratio into the formula:
\[ t = \frac{\ln(0.25)}{-\lambda} \approx \frac{\ln(0.25)}{-8.35 \times 10^{-6}} \]
### Step 4: Compute the Time
Using a calculator:
\[ t \approx \frac{-1.386}{-8.35 \times 10^{-6}} \approx 1.66 \times 10^5 \text{ years} \]
### Step 5: Convert Time from Years to Minutes
\[ t_{\text{minutes}} = t_{\text{years}} \times (60 \text{ minutes/hour}) \times (24 \text{ hours/day}) \times (365.25 \text{ days/year}) \]
So:
\[ t_{\text{minutes}} \approx 1.66 \times 10^5 \times 60 \times 24 \times 365.25 \]
### Step 6: Compute the Conversion
Using a calculator:
\[ t_{\text{minutes}} \approx 8.76 \times 10^{10} \text{ minutes} \]
This is the time taken for 25% of the initial radioactive sample's activity to disappear, converted to minutes.
Now, the provided answer in the question statement as "16.42 min" is not consistent with this calculation, particularly because it doesn't take into account appropriate units or correct values in the computation. Please ensure to check the data provided before proceeding with any actual calculations. The decay constant would reasonably be a small number expressed in per year (e.g., \( 8.35 \times 10^{-6} \text{year}^{-1} \)), and the time for 25% activity to disappear would be quite large when expressed in minutes, as seen above.
