Two identical samples (same material and same amout) P and Q of a radioactive substance having mean life T are observed to have activities `A_P` and `A_Q` respectively at the time of observation. If P is older than Q, then the difference in their age is

Options

A. (a) `T 1n ((A_P)/(A_Q))`
B. (b) `T 1n ((A_Q)/(A_P))`
C. (c) `T((A_P)/(A_Q))`
D. (d) `T((A_Q)/(A_P))`

Explanation

To find the difference in age between two identical samples P and Q of a radioactive substance, we can follow these steps:

### Step 1: Understand the relationship between activity and time
The activity \( A \) of a radioactive sample is given by the equation:
\[ A = A_0 e^{-\lambda t} \]
where:
- \( A_0 \) is the initial activity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 2: Write the activity equations for both samples
For sample P (which is older), the activity at time \( t_1 \) can be expressed as:
\[ A_P = A_0 e^{-\lambda t_1} \]
For sample Q (which is younger), the activity at time \( t_2 \) can be expressed as:
\[ A_Q = A_0 e^{-\lambda t_2} \]

### Step 3: Relate the mean life to the decay constant
The mean life \( T \) of a radioactive substance is related to the decay constant \( \lambda \) by:
\[ T = \frac{1}{\lambda} \]
Thus, we can express \( \lambda \) as:
\[ \lambda = \frac{1}{T} \]

### Step 4: Express the ages in terms of activities
From the activity equations, we can take the natural logarithm to express the ages:
For sample P:
\[ t_1 = -\frac{1}{\lambda} \ln \left( \frac{A_P}{A_0} \right) \]
For sample Q:
\[ t_2 = -\frac{1}{\lambda} \ln \left( \frac{A_Q}{A_0} \right) \]

### Step 5: Calculate the difference in age
Since sample P is older than sample Q, we find the difference in their ages:
\[ \Delta t = t_1 - t_2 \]
Substituting the expressions for \( t_1 \) and \( t_2 \):
\[ \Delta t = -\frac{1}{\lambda} \ln \left( \frac{A_P}{A_0} \right) + \frac{1}{\lambda} \ln \left( \frac{A_Q}{A_0} \right) \]
This simplifies to:
\[ \Delta t = \frac{1}{\lambda} \left( \ln \left( \frac{A_Q}{A_0} \right) - \ln \left( \frac{A_P}{A_0} \right) \right) \]
Using the properties of logarithms, this can be rewritten as:
\[ \Delta t = \frac{1}{\lambda} \ln \left( \frac{A_Q}{A_P} \right) \]

### Step 6: Substitute \( \lambda \) with \( \frac{1}{T} \)
Substituting \( \lambda \) back in terms of mean life \( T \):
\[ \Delta t = T \ln \left( \frac{A_Q}{A_P} \right) \]

### Final Answer
Thus, the difference in their ages is:
\[ \Delta t = T \ln \left( \frac{A_Q}{A_P} \right) \]