A radioactive material decays by simultaneous emission of two particle from the with respective half - lives `1620` and `810` year . The time , in year , after which one - fourth of the material remains is

Options

A. 1080 years`
B. 2340 years
C. 4860 years
D. 3240 years

Answer

To solve the problem of determining the time after which one-fourth of a radioactive material remains, we need to consider the simultaneous decay of two particles with given half-lives. Here’s a step-by-step solution:

### Step 1: Understand the half-lives
We have two half-lives:
- Half-life of the first particle, \( t_{1/2,1} = 1620 \) years
- Half-life of the second particle, \( t_{1/2,2} = 810 \) years

### Step 2: Calculate the decay constants
The decay constant \( \lambda \) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
For the first particle:
\[ \lambda_1 = \frac{\ln(2)}{1620} \]
For the second particle:
\[ \lambda_2 = \frac{\ln(2)}{810} \]

### Step 3: Find the total decay constant
Since the particles decay simultaneously, the total decay constant \( \lambda_{total} \) is the sum of the individual decay constants:
\[ \lambda_{total} = \lambda_1 + \lambda_2 = \frac{\ln(2)}{1620} + \frac{\ln(2)}{810} \]
Factoring out \( \ln(2) \):
\[ \lambda_{total} = \ln(2) \left( \frac{1}{1620} + \frac{1}{810} \right) \]
Finding a common denominator (which is 1620):
\[ \frac{1}{810} = \frac{2}{1620} \]
Thus,
\[ \lambda_{total} = \ln(2) \left( \frac{1 + 2}{1620} \right) = \ln(2) \left( \frac{3}{1620} \right) = \frac{3 \ln(2)}{1620} \]

### Step 4: Calculate the effective half-life of the mixture
The effective half-life \( t_{1/2, total} \) can be calculated using:
\[ t_{1/2, total} = \frac{\ln(2)}{\lambda_{total}} = \frac{\ln(2)}{\frac{3 \ln(2)}{1620}} = \frac{1620}{3} = 540 \text{ years} \]

### Step 5: Determine the time for one-fourth of the material to remain
To find the time after which one-fourth of the material remains, we note that it takes two half-lives to reduce the material to one-fourth:
\[ t = 2 \times t_{1/2, total} = 2 \times 540 = 1080 \text{ years} \]

### Final Answer
The time after which one-fourth of the material remains is **1080 years**.

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