Half-lives of two radioactive substances `A` and `B` are respectively `20` minutes and `40` minutes. Initially, he sample of `A` and `B` have equal number of nuclei. After `80` minutes the ratio of the remaining number of `A` and `B` nuclei is :

Options

A. `1 : 16`
B. `4 : 1`
C. `1 : 4`
D. `1 : 1`

Explanation

To solve the problem, we need to determine the remaining number of nuclei of substances A and B after 80 minutes, given their half-lives.

### Step 1: Understanding Half-Life
The half-life of a radioactive substance is the time required for half of the radioactive nuclei in a sample to decay.

- For substance A, the half-life is 20 minutes.
- For substance B, the half-life is 40 minutes.

### Step 2: Calculate the Number of Half-Lives
Next, we need to determine how many half-lives have passed for each substance after 80 minutes.

- For substance A:
\[ \text{Number of half-lives} = \frac{80 \text{ minutes}}{20 \text{ minutes}} = 4 \]

- For substance B:
\[ \text{Number of half-lives} = \frac{80 \text{ minutes}}{40 \text{ minutes}} = 2 \]

### Step 3: Calculate Remaining Nuclei
Let the initial number of nuclei of both substances be \( N_0 \).

- For substance A after 4 half-lives:
\[ N_A = N_0 \left(\frac{1}{2}\right)^4 = N_0 \left(\frac{1}{16}\right) \]

- For substance B after 2 half-lives:
\[ N_B = N_0 \left(\frac{1}{2}\right)^2 = N_0 \left(\frac{1}{4}\right) \]

### Step 4: Calculate the Ratio of Remaining Nuclei
Now, we can find the ratio of the remaining nuclei of A to B:
\[ \text{Ratio} = \frac{N_A}{N_B} = \frac{N_0 \left(\frac{1}{16}\right)}{N_0 \left(\frac{1}{4}\right)} = \frac{\frac{1}{16}}{\frac{1}{4}} = \frac{1}{16} \times \frac{4}{1} = \frac{4}{16} = \frac{1}{4} \]

### Final Answer
The ratio of the remaining number of nuclei of A to B after 80 minutes is \( \frac{1}{4} \).

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