Two radioactive substances have half-lives T and 2T. Initially, they have equal number of nuclei. After time `t=4T`, the ratio of their number of nuclei is x and the ratio of their activity is y. Then,

Options

A. (a) `x=1//8`
B. (b) `x=1//4`
C. (c) `y=1//2`
D. (d) `y=1//4`

Explanation

To solve the problem, we need to find the ratio of the number of nuclei and the ratio of the activities of two radioactive substances with different half-lives after a certain time has passed. Let's break down the solution step by step.

### Step 1: Understand the half-lives and initial conditions
We have two radioactive substances:
- Substance 1 has a half-life of \( T \).
- Substance 2 has a half-life of \( 2T \).

Initially, both substances have the same number of nuclei, denoted as \( N_0 \).

### Step 2: Calculate the number of half-lives elapsed
We are given that the time elapsed is \( t = 4T \).

For Substance 1:
- The number of half-lives \( n_1 \) is calculated as:
\[ n_1 = \frac{t}{T} = \frac{4T}{T} = 4 \]

For Substance 2:
- The number of half-lives \( n_2 \) is calculated as:
\[ n_2 = \frac{t}{2T} = \frac{4T}{2T} = 2 \]

### Step 3: Calculate the remaining number of nuclei
The remaining number of nuclei for each substance can be calculated using the formula:
\[ N = N_0 \left( \frac{1}{2} \right)^n \]

For Substance 1:
\[ N_1 = N_0 \left( \frac{1}{2} \right)^4 = N_0 \cdot \frac{1}{16} \]

For Substance 2:
\[ N_2 = N_0 \left( \frac{1}{2} \right)^2 = N_0 \cdot \frac{1}{4} \]

### Step 4: Find the ratio of the number of nuclei
Now we can find the ratio \( X \) of the number of nuclei of Substance 1 to Substance 2:
\[ X = \frac{N_1}{N_2} = \frac{N_0 \cdot \frac{1}{16}}{N_0 \cdot \frac{1}{4}} = \frac{\frac{1}{16}}{\frac{1}{4}} = \frac{1}{16} \cdot \frac{4}{1} = \frac{1}{4} \]

### Step 5: Calculate the activities
The activity \( A \) of a radioactive substance is given by:
\[ A = \lambda N \]
where \( \lambda \) is the decay constant. The decay constant is related to the half-life \( T \) by:
\[ \lambda = \frac{\ln(2)}{T} \]

For Substance 1:
\[ A_1 = \lambda_1 N_1 = \frac{\ln(2)}{T} \cdot N_0 \cdot \frac{1}{16} \]

For Substance 2:
\[ A_2 = \lambda_2 N_2 = \frac{\ln(2)}{2T} \cdot N_0 \cdot \frac{1}{4} \]

### Step 6: Find the ratio of the activities
Now we can find the ratio \( Y \) of the activities:
\[ Y = \frac{A_1}{A_2} = \frac{\frac{\ln(2)}{T} \cdot N_0 \cdot \frac{1}{16}}{\frac{\ln(2)}{2T} \cdot N_0 \cdot \frac{1}{4}} \]

The \( N_0 \) and \( \ln(2) \) terms cancel out:
\[ Y = \frac{\frac{1}{16}}{\frac{1}{8}} = \frac{1}{16} \cdot \frac{8}{1} = \frac{1}{2} \]

### Final Results
- The ratio of the number of nuclei \( X = \frac{1}{4} \).
- The ratio of the activities \( Y = \frac{1}{2} \).