A radioactive sample is undergoing `alpha` - decay. At time `t_1`, its activity is A & at A another time `t_2`, the activity is `A/5`. What is the average life time for the sample

Options

A. `(t_1-t_2)/ ln2`
B. `(t_1-t_2) ln5`
C. `(t_1-t_2)/2`
D. `(t_2-t_1)/ ln5`

Answer

To solve the problem of finding the average lifetime of a radioactive sample undergoing alpha decay, we can follow these steps:

### Step 1: Understand the relationship between activity and decay
The activity \( A \) of a radioactive sample is proportional to the number of undecayed nuclei \( N \) in the sample. The relationship can be expressed as:
\[ A = \lambda N \]
where \( \lambda \) is the decay constant.

### Step 2: Set up the equations for the two time points
At time \( t_1 \), the activity is \( A \):
\[ A = \lambda N_0 e^{-\lambda t_1} \]
where \( N_0 \) is the initial number of nuclei.

At time \( t_2 \), the activity is \( \frac{A}{5} \):
\[ \frac{A}{5} = \lambda N_0 e^{-\lambda t_2} \]

### Step 3: Substitute the first equation into the second
From the first equation, we can express \( A \) in terms of \( N_0 \):
\[ N_0 = \frac{A}{\lambda e^{-\lambda t_1}} \]
Now substitute this into the second equation:
\[ \frac{A}{5} = \lambda \left(\frac{A}{\lambda e^{-\lambda t_1}}\right) e^{-\lambda t_2} \]
This simplifies to:
\[ \frac{A}{5} = \frac{A e^{-\lambda t_2}}{5 e^{-\lambda t_1}} \]

### Step 4: Cancel out \( A \) from both sides
Since \( A \) is common on both sides, we can cancel it:
\[ \frac{1}{5} = \frac{e^{-\lambda t_2}}{e^{-\lambda t_1}} \]
This can be rewritten as:
\[ 5 = e^{-\lambda t_2 + \lambda t_1} \]
or
\[ 5 = e^{\lambda(t_1 - t_2)} \]

### Step 5: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \ln(5) = \lambda(t_1 - t_2) \]

### Step 6: Solve for the decay constant \( \lambda \)
Rearranging the equation gives:
\[ \lambda = \frac{\ln(5)}{t_2 - t_1} \]

### Step 7: Calculate the average lifetime \( \tau \)
The average lifetime \( \tau \) is given by:
\[ \tau = \frac{1}{\lambda} \]
Substituting the expression for \( \lambda \):
\[ \tau = \frac{t_2 - t_1}{\ln(5)} \]

### Final Answer
Thus, the average lifetime of the sample is:
\[ \tau = \frac{t_2 - t_1}{\ln(5)} \]

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