A radioactive sample contains two different types of radioactive nuclei. A-with half-time 5 days and B-type with half life 30 days. Initially the decay rate of A-type nuclei is 64 times that of B type of nuclei . Their decay rates will be equal when time is 9n days. Find the value of n.

Answer

To solve the problem, we need to find the value of \( n \) such that the decay rates of two different types of radioactive nuclei, A and B, become equal after a certain time. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the given data
- A-type nuclei have a half-life (\( T_{1/2} \)) of 5 days.
- B-type nuclei have a half-life (\( T_{1/2} \)) of 30 days.
- Initially, the decay rate of A-type nuclei is 64 times that of B-type nuclei.

### Step 2: Define the decay constants
The decay constant (\( \lambda \)) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
For A-type:
\[ \lambda_A = \frac{\ln(2)}{5} \]
For B-type:
\[ \lambda_B = \frac{\ln(2)}{30} \]

### Step 3: Write the expressions for the number of nuclei remaining
Let \( N_{0A} \) be the initial number of A-type nuclei and \( N_{0B} \) be the initial number of B-type nuclei. Given that \( N_{0A} = 64 N_{0B} \).

The number of A-type nuclei remaining after time \( t \) is:
\[ N_A(t) = N_{0A} e^{-\lambda_A t} \]
The number of B-type nuclei remaining after time \( t \) is:
\[ N_B(t) = N_{0B} e^{-\lambda_B t} \]

### Step 4: Set the equations equal for decay rates
We want to find when the decay rates (or activities) of A and B become equal:
\[ N_A(9n) = N_B(9n) \]
Substituting the expressions:
\[ N_{0A} e^{-\lambda_A (9n)} = N_{0B} e^{-\lambda_B (9n)} \]
Substituting \( N_{0A} = 64 N_{0B} \):
\[ 64 N_{0B} e^{-\lambda_A (9n)} = N_{0B} e^{-\lambda_B (9n)} \]
Dividing both sides by \( N_{0B} \):
\[ 64 e^{-\lambda_A (9n)} = e^{-\lambda_B (9n)} \]

### Step 5: Rearranging the equation
Taking the natural logarithm of both sides:
\[ \ln(64) - \lambda_A (9n) = -\lambda_B (9n) \]
This simplifies to:
\[ \ln(64) = \lambda_A (9n) - \lambda_B (9n) \]
Factoring out \( 9n \):
\[ \ln(64) = ( \lambda_A - \lambda_B )(9n) \]

### Step 6: Substitute the values of \( \lambda_A \) and \( \lambda_B \)
Substituting the decay constants:
\[ \ln(64) = \left( \frac{\ln(2)}{5} - \frac{\ln(2)}{30} \right)(9n) \]
Finding a common denominator (30):
\[ \ln(64) = \left( \frac{6\ln(2)}{30} - \frac{\ln(2)}{30} \right)(9n) \]
This simplifies to:
\[ \ln(64) = \left( \frac{5\ln(2)}{30} \right)(9n) \]

### Step 7: Solve for \( n \)
Now, substituting \( \ln(64) = 6 \ln(2) \):
\[ 6 \ln(2) = \left( \frac{5\ln(2)}{30} \right)(9n) \]
Dividing both sides by \( \ln(2) \):
\[ 6 = \frac{5}{30} (9n) \]
This simplifies to:
\[ 6 = \frac{3}{2} n \]
Multiplying both sides by \( \frac{2}{3} \):
\[ n = 4 \]

### Final Answer
Thus, the value of \( n \) is \( 4 \).

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