The mean lives of an unstable nucleus in two different decay processes are 1620 yr and 405 yr, respectively. Find out the time during which three-fourth of a sample will decay.

Explanation

To solve the problem of finding the time during which three-fourths of a sample will decay, we can follow these steps:

### Step 1: Understand the Mean Life and Decay Constant
The mean life \( t \) of a nucleus is related to the decay constant \( \lambda \) by the formula:
\[ \lambda = \frac{1}{t} \]
Given the mean lives:
- \( t_1 = 1620 \) years
- \( t_2 = 405 \) years

### Step 2: Calculate the Decay Constants
Using the formula for decay constant:
\[ \lambda_1 = \frac{1}{t_1} = \frac{1}{1620} \, \text{yr}^{-1} \]
\[ \lambda_2 = \frac{1}{t_2} = \frac{1}{405} \, \text{yr}^{-1} \]

### Step 3: Find the Total Decay Constant
The total decay constant \( \lambda \) for both processes is the sum of the individual decay constants:
\[ \lambda = \lambda_1 + \lambda_2 = \frac{1}{1620} + \frac{1}{405} \]
To add these fractions, we need a common denominator. The least common multiple of 1620 and 405 is 1620. Thus:
\[ \lambda = \frac{1}{1620} + \frac{4}{1620} = \frac{5}{1620} = \frac{1}{324} \, \text{yr}^{-1} \]

### Step 4: Set Up the Decay Equation
When three-fourths of the sample has decayed, one-fourth remains. If \( N_0 \) is the initial number of nuclei, then:
\[ N = \frac{N_0}{4} \]
Using the decay formula:
\[ N = N_0 e^{-\lambda t} \]
Substituting the values:
\[ \frac{N_0}{4} = N_0 e^{-\lambda t} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{4} = e^{-\lambda t} \]

### Step 5: Solve for Time \( t \)
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{4}\right) = -\lambda t \]
Substituting \( \lambda = \frac{1}{324} \):
\[ \ln\left(\frac{1}{4}\right) = -\frac{1}{324} t \]
We know that \( \ln\left(\frac{1}{4}\right) = -\ln(4) = -2\ln(2) \):
\[ -2\ln(2) = -\frac{1}{324} t \]
Thus:
\[ t = 2 \ln(2) \times 324 \]
Calculating \( 2 \ln(2) \):
\[ \ln(2) \approx 0.693 \]
So:
\[ t \approx 2 \times 0.693 \times 324 \approx 449 \, \text{years} \]

### Final Answer
The time during which three-fourths of the sample will decay is approximately **449 years**.

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