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. (a) `(T)/(1n2)1n(2A_1)/(A_2)`
B. (b) `T1n(A_1)/(A_2)`
C. (c) `(T)/(1n2)1n(A_2)/(2A_1)`
D. (d) `T1n(A_2)/(2A_1)`

Explanation

To solve the problem step by step, we need to analyze the relationship between the activities of two samples of radioactive material kept at different times and their respective ages.

### Step 1: Understand the relationship between activity and time
The activity \( A \) of a radioactive substance is related to its initial activity \( A_0 \) and the decay constant \( \lambda \) by the equation:
\[ A = A_0 e^{-\lambda t} \]
where \( t \) is the time elapsed since the material was kept.

### Step 2: Set up the equations for both samples
Let’s denote:
- For the first sample, the initial activity is \( A_{01} \) and the time elapsed is \( t_1 \).
- For the second sample, the initial activity is \( A_{02} = 2 A_{01} \) (since the second quantity is twice the first) and the time elapsed is \( t_2 \).

From the activity equation, we can write:
\[ A_1 = A_{01} e^{-\lambda t_1} \]
\[ A_2 = A_{02} e^{-\lambda t_2} = 2 A_{01} e^{-\lambda t_2} \]

### Step 3: Relate the activities
Now we can express the ratio of the activities:
\[ \frac{A_1}{A_2} = \frac{A_{01} e^{-\lambda t_1}}{2 A_{01} e^{-\lambda t_2}} = \frac{e^{-\lambda t_1}}{2 e^{-\lambda t_2}} \]
This simplifies to:
\[ \frac{A_1}{A_2} = \frac{1}{2} e^{\lambda (t_2 - t_1)} \]

### Step 4: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{A_1}{A_2}\right) = \ln\left(\frac{1}{2}\right) + \lambda (t_2 - t_1) \]

### Step 5: Solve for the time difference
Rearranging the equation to isolate \( t_2 - t_1 \):
\[ \lambda (t_2 - t_1) = \ln\left(\frac{A_1}{A_2}\right) - \ln\left(\frac{1}{2}\right) \]
Using properties of logarithms, we can express this as:
\[ t_2 - t_1 = \frac{1}{\lambda} \left( \ln\left(\frac{A_1}{A_2}\right) + \ln(2) \right) \]
This can be simplified to:
\[ t_2 - t_1 = \frac{1}{\lambda} \ln\left(\frac{2 A_1}{A_2}\right) \]

### Step 6: Substitute decay constant
The decay constant \( \lambda \) can be expressed in terms of the half-life \( T \):
\[ \lambda = \frac{\ln(2)}{T} \]
Substituting this into the equation gives:
\[ t_2 - t_1 = \frac{T}{\ln(2)} \ln\left(\frac{2 A_1}{A_2}\right) \]

### Final Result
Thus, the age difference between the two samples is:
\[ t_2 - t_1 = \frac{T}{\ln(2)} \ln\left(\frac{2 A_1}{A_2}\right) \]