At any instant, the ratio of the amounts of two radioactive substance is `2:1`. If their half-lives be, respectively, `12h` and `16h`, then after two days, what will be the ratio of the substances?

Options

A. `1 : 1`
B. `2 : 1`
C. `1 : 2`
D. `1 : 4`

Explanation

To solve the problem step by step, we will first define the variables and then calculate the amounts of the two radioactive substances after two days.

### Step 1: Define the initial amounts and half-lives
Let the initial amounts of the two radioactive substances be:
- Substance A: \( N_{0A} = 2x \)
- Substance B: \( N_{0B} = x \)

The half-lives of the substances are:
- Half-life of Substance A (\( t_{1/2A} \)) = 12 hours
- Half-life of Substance B (\( t_{1/2B} \)) = 16 hours

### Step 2: Calculate the number of half-lives in two days
Two days is equivalent to 48 hours. We can calculate the number of half-lives for each substance:
- For Substance A:
\[ N_{1} = \frac{48 \text{ hours}}{12 \text{ hours}} = 4 \text{ half-lives} \]
- For Substance B:
\[ N_{2} = \frac{48 \text{ hours}}{16 \text{ hours}} = 3 \text{ half-lives} \]

### Step 3: Calculate the remaining amounts after two days
Using the formula for radioactive decay:
\[ N = N_0 \left( \frac{1}{2} \right)^N \]
we can find the remaining amounts after 4 and 3 half-lives.

- Remaining amount of Substance A:
\[ N_A = N_{0A} \left( \frac{1}{2} \right)^{N_{1}} = 2x \left( \frac{1}{2} \right)^{4} = 2x \cdot \frac{1}{16} = \frac{x}{8} \]

- Remaining amount of Substance B:
\[ N_B = N_{0B} \left( \frac{1}{2} \right)^{N_{2}} = x \left( \frac{1}{2} \right)^{3} = x \cdot \frac{1}{8} = \frac{x}{8} \]

### Step 4: Calculate the final ratio of the substances
Now we can find the ratio of the remaining amounts of the two substances:
\[ \text{Ratio} = \frac{N_A}{N_B} = \frac{\frac{x}{8}}{\frac{x}{8}} = 1 \]

Thus, the ratio of the amounts of the two substances after two days is:
\[ \text{Ratio} = 1:1 \]

### Final Answer
The ratio of the two radioactive substances after two days is \( 1:1 \).

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