Nuclei X and Y convert into a stable nucleus Z. At t=0 , the number of nuclei of X is 8 times that of Y. Half life of X is 1 hour and half life of y=2 hour. Find the time (in hour) at which rate of disintegration of X and Y are equal.

Answer

To solve the problem, we need to find the time at which the rate of disintegration of nuclei X and Y becomes equal. Let's break down the solution step-by-step.

### Step 1: Define the Initial Conditions
At time \( t = 0 \):
- Let the number of nuclei of Y be \( N_Y(0) = n_0 \).
- Then, the number of nuclei of X will be \( N_X(0) = 8n_0 \) (since X is 8 times Y).

### Step 2: Write the Decay Formulas
The number of nuclei remaining after time \( t \) can be expressed using the half-life formula:
\[ N(t) = N(0) \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
where \( T_{1/2} \) is the half-life of the substance.

For nucleus X (half-life = 1 hour):
\[ N_X(t) = 8n_0 \left( \frac{1}{2} \right)^{\frac{t}{1}} = 8n_0 \left( \frac{1}{2} \right)^{t} \]

For nucleus Y (half-life = 2 hours):
\[ N_Y(t) = n_0 \left( \frac{1}{2} \right)^{\frac{t}{2}} = n_0 \left( \frac{1}{2} \right)^{\frac{t}{2}} \]

### Step 3: Determine the Rates of Disintegration
The rate of disintegration (activity) of a radioactive substance is given by:
\[ R = \lambda N \]
where \( \lambda \) is the decay constant related to half-life by:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]

For nucleus X:
\[ \lambda_X = \frac{\ln(2)}{1} = \ln(2) \]
Thus, the rate of disintegration of X is:
\[ R_X(t) = \lambda_X N_X(t) = \ln(2) \cdot 8n_0 \left( \frac{1}{2} \right)^{t} \]

For nucleus Y:
\[ \lambda_Y = \frac{\ln(2)}{2} = \frac{\ln(2)}{2} \]
Thus, the rate of disintegration of Y is:
\[ R_Y(t) = \lambda_Y N_Y(t) = \frac{\ln(2)}{2} \cdot n_0 \left( \frac{1}{2} \right)^{\frac{t}{2}} \]

### Step 4: Set the Rates Equal
To find the time when the rates are equal:
\[ R_X(t) = R_Y(t) \]
Substituting the expressions we derived:
\[ \ln(2) \cdot 8n_0 \left( \frac{1}{2} \right)^{t} = \frac{\ln(2)}{2} \cdot n_0 \left( \frac{1}{2} \right)^{\frac{t}{2}} \]

### Step 5: Simplify the Equation
Cancel \( n_0 \) and \( \ln(2) \) from both sides:
\[ 8 \left( \frac{1}{2} \right)^{t} = \frac{1}{2} \left( \frac{1}{2} \right)^{\frac{t}{2}} \]
Multiply both sides by 2:
\[ 16 \left( \frac{1}{2} \right)^{t} = \left( \frac{1}{2} \right)^{\frac{t}{2}} \]

### Step 6: Take Logarithms
Taking logarithms on both sides:
\[ \ln(16) - t \ln(2) = -\frac{t}{2} \ln(2) \]
Rearranging gives:
\[ \ln(16) = t \ln(2) \left( 1 - \frac{1}{2} \right) \]
\[ \ln(16) = \frac{t}{2} \ln(2) \]

### Step 7: Solve for \( t \)
\[ t = \frac{2 \ln(16)}{\ln(2)} = \frac{2 \cdot 4 \ln(2)}{\ln(2)} = 8 \text{ hours} \]

### Final Answer
The time at which the rate of disintegration of X and Y are equal is:
\[ \boxed{8 \text{ hours}} \]