A radioactive material decays by simulataneous 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
B. 2430
C. 3240
D. 4860

Answer

To solve the problem of determining the time after which one-fourth of a radioactive material remains, given the half-lives of two particles, we can follow these steps:

### Step 1: Identify the half-lives
The half-lives provided are:
- \( T_1 = 1620 \) years (for the first particle)
- \( T_2 = 810 \) years (for the second particle)

### 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:
\[ \lambda_1 = \frac{\ln(2)}{T_1} = \frac{\ln(2)}{1620} \]
For the second particle:
\[ \lambda_2 = \frac{\ln(2)}{T_2} = \frac{\ln(2)}{810} \]

### Step 3: Calculate the effective decay constant
Since the two particles decay simultaneously, the effective decay constant \( \lambda_0 \) is the sum of the individual decay constants:
\[ \lambda_0 = \lambda_1 + \lambda_2 = \frac{\ln(2)}{1620} + \frac{\ln(2)}{810} \]
Factoring out \( \ln(2) \):
\[ \lambda_0 = \ln(2) \left( \frac{1}{1620} + \frac{1}{810} \right) \]

### Step 4: Simplify the expression
To combine the fractions:
\[ \frac{1}{1620} + \frac{1}{810} = \frac{1}{1620} + \frac{2}{1620} = \frac{3}{1620} = \frac{1}{540} \]
Thus, we have:
\[ \lambda_0 = \ln(2) \cdot \frac{1}{540} \]

### Step 5: Calculate the effective half-life
The effective half-life \( T_0 \) can be found using:
\[ T_0 = \frac{\ln(2)}{\lambda_0} = \frac{\ln(2)}{\ln(2) \cdot \frac{1}{540}} = 540 \text{ years} \]

### Step 6: Determine the number of half-lives for one-fourth remaining
To find the time \( t \) after which one-fourth of the material remains, we can use the relationship:
\[ \frac{N}{N_0} = \left( \frac{1}{2} \right)^n \]
For one-fourth remaining, \( \frac{N}{N_0} = \frac{1}{4} = \left( \frac{1}{2} \right)^2 \), which means \( n = 2 \).

### Step 7: Relate the number of half-lives to time
The number of half-lives \( n \) is also given by:
\[ n = \frac{t}{T_0} \]
Substituting the values:
\[ 2 = \frac{t}{540} \]
Thus, solving for \( t \):
\[ t = 2 \times 540 = 1080 \text{ years} \]

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

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