A radioactive material of half-life `T` was kept in a nuclear reactor at two different instants. The quantity kept second time was twice of the kept first time. If now their present activities are `A_1` and `A_2` respectively, then their age difference equals

Options

A. `(T)/(ln2)|ln.(2A_(1))/(A_(2))|`
B. `T|ln.(2A_(1))/(A_(2))|`
C. `(T)/(ln2)|ln.(A_(1))/(2A_(2))|`
D. `T|ln.(A_(2))/(2A_(1))|`

Answer

To solve the problem step by step, we will analyze the given information about the radioactive material and its activities at two different times.

### Step 1: Understand the problem
We have a radioactive material with a half-life \( T \). The first quantity of material is kept at time \( t_1 \) and the second quantity, which is twice the first, is kept at time \( t_2 \). We need to find the age difference between the two quantities based on their present activities \( A_1 \) and \( A_2 \).

### Step 2: Define initial activities
Let:
- The initial activity of the first quantity be \( A_0 \).
- The initial activity of the second quantity be \( A_0' = 2A_0 \) (since it is twice the first).

### Step 3: Write the equations for activities
The activity of a radioactive substance at any time \( t \) is given by:
\[ A = A_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

For the first quantity at time \( t_1 \):
\[ A_1 = A_0 e^{-\lambda t_1} \]

For the second quantity at time \( t_2 \):
\[ A_2 = A_0' e^{-\lambda t_2} = 2A_0 e^{-\lambda t_2} \]

### Step 4: Rearranging the equations
From the first equation, we can express \( A_0 \):
\[ A_0 = \frac{A_1}{e^{-\lambda t_1}} = A_1 e^{\lambda t_1} \]

From the second equation:
\[ A_0' = \frac{A_2}{e^{-\lambda t_2}} = \frac{A_2}{2A_0 e^{-\lambda t_2}} = \frac{A_2}{2A_1 e^{\lambda t_1}} e^{\lambda t_2} \]

### Step 5: Find expressions for \( t_1 \) and \( t_2 \)
From the equations for \( A_1 \) and \( A_2 \):
1. For \( A_1 \):
\[ A_1 = A_0 e^{-\lambda t_1} \implies t_1 = \frac{1}{\lambda} \ln\left(\frac{A_0}{A_1}\right) \]

2. For \( A_2 \):
\[ A_2 = 2A_0 e^{-\lambda t_2} \implies t_2 = \frac{1}{\lambda} \ln\left(\frac{2A_0}{A_2}\right) \]

### Step 6: Calculate the age difference
The age difference \( \Delta t = t_1 - t_2 \):
\[ \Delta t = \frac{1}{\lambda} \left( \ln\left(\frac{A_0}{A_1}\right) - \ln\left(\frac{2A_0}{A_2}\right) \right) \]
Using properties of logarithms:
\[ \Delta t = \frac{1}{\lambda} \left( \ln\left(\frac{A_0}{A_1}\right) - \left( \ln(2) + \ln\left(\frac{A_0}{A_2}\right) \right) \right) \]
This simplifies to:
\[ \Delta t = \frac{1}{\lambda} \left( \ln\left(\frac{A_0 A_2}{2 A_1}\right) \right) \]

### Step 7: Substitute \( \lambda \) in terms of half-life
Recall that \( \lambda = \frac{\ln(2)}{T} \):
\[ \Delta t = \frac{T}{\ln(2)} \ln\left(\frac{A_2}{2 A_1}\right) \]

### Final Answer
Thus, the age difference between the two quantities of radioactive material is:
\[ \Delta t = \frac{T}{\ln(2)} \ln\left(\frac{A_2}{2 A_1}\right) \]