Two radioactive samples of different elements (half-lives `t_1` and `t_2` respectively) have same number of nuclei at `t=0`. The time after which their activities are same is

Options

A. (a) `(t_1t_2)/(0.693(t_2-t_1))1n(t_2)/(t_1)`
B. (b) `(t_1t_2)/(0.693)1n(t_2)/(t_1)`
C. (c) `(t_1t_2)/(0.693(t_1+t_2))1n(t_2)/(t_1)`
D. (d) None of these

Explanation

To solve the problem of finding the time after which the activities of two radioactive samples become equal, we can follow these steps:

### Step 1: Understand the relationship between activity and decay constant
The activity \( R \) of a radioactive sample is given by the formula:
\[ R = \lambda N \]
where \( \lambda \) is the decay constant and \( N \) is the number of undecayed nuclei.

### Step 2: Use the radioactive decay law
The number of undecayed nuclei at time \( t \) is given by:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei.

### Step 3: Write the expressions for the activities of both samples
Let the two samples have half-lives \( t_1 \) and \( t_2 \), and decay constants \( \lambda_1 \) and \( \lambda_2 \) respectively. The decay constants can be related to half-lives as follows:
\[ \lambda_1 = \frac{\ln 2}{t_1}, \quad \lambda_2 = \frac{\ln 2}{t_2} \]
Thus, the activities at time \( t \) for both samples can be expressed as:
\[ R_1(t) = \lambda_1 N_0 e^{-\lambda_1 t} \]
\[ R_2(t) = \lambda_2 N_0 e^{-\lambda_2 t} \]

### Step 4: Set the activities equal to each other
To find the time \( t \) when the activities are the same, we set \( R_1(t) = R_2(t) \):
\[ \lambda_1 N_0 e^{-\lambda_1 t} = \lambda_2 N_0 e^{-\lambda_2 t} \]
Since \( N_0 \) is the same for both samples, we can cancel it out:
\[ \lambda_1 e^{-\lambda_1 t} = \lambda_2 e^{-\lambda_2 t} \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \frac{\lambda_1}{\lambda_2} = e^{(\lambda_1 - \lambda_2)t} \]

### Step 6: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln\left(\frac{\lambda_1}{\lambda_2}\right) = (\lambda_1 - \lambda_2)t \]

### Step 7: Solve for \( t \)
Now, solve for \( t \):
\[ t = \frac{\ln\left(\frac{\lambda_1}{\lambda_2}\right)}{\lambda_1 - \lambda_2} \]

### Step 8: Substitute the decay constants
Substituting \( \lambda_1 \) and \( \lambda_2 \):
\[ t = \frac{\ln\left(\frac{\frac{\ln 2}{t_1}}{\frac{\ln 2}{t_2}}\right)}{\frac{\ln 2}{t_1} - \frac{\ln 2}{t_2}} \]
This simplifies to:
\[ t = \frac{\ln\left(\frac{t_2}{t_1}\right)}{\frac{\ln 2}{t_1} - \frac{\ln 2}{t_2}} \]

### Step 9: Final expression
This can be further simplified to:
\[ t = \frac{t_1 t_2}{\ln 2} \cdot \frac{\ln\left(\frac{t_2}{t_1}\right)}{t_2 - t_1} \]

### Conclusion
Thus, the time after which the activities of the two radioactive samples become equal is given by:
\[ t = \frac{t_1 t_2}{\ln 2} \cdot \frac{\ln\left(\frac{t_2}{t_1}\right)}{t_2 - t_1} \]