For a certain reaction `Ato` products, the `t_(1//2)` as a function of `[A]_0` is given as below:
`{:([A]_0(M),0.1,0.025),(t_(1//2),100,50):}`
Which of the following is true :

Options

A. The order of `1/2`
B. `t_(1//2)`would be `100sqrt10` min for `[A]_0=1` M
C. The order is 1
D. `t_(1//2)`would be `100` min for `[A]_0=1` M

Explanation

To solve the problem, we need to analyze the relationship between the half-life \( t_{1/2} \) and the initial concentration \( [A]_0 \) of the reactant \( A \). The relationship is given by the formula:

\[ t_{1/2} \propto [A]_0^{(1-n)} \]

where \( n \) is the order of the reaction.

### Step-by-Step Solution:

1. **Set Up the Proportionality**:
We have two sets of data:
- For \( [A]_0 = 0.1 \, M \), \( t_{1/2} = 100 \, \text{minutes} \)
- For \( [A]_0 = 0.025 \, M \), \( t_{1/2} = 50 \, \text{minutes} \)

Using the proportionality, we can write:
\[ t_{1/2} \propto [A]_0^{(1-n)} \]

2. **Write the Equations**:
From the first data point:
\[ 100 \propto (0.1)^{(1-n)} \]
From the second data point:
\[ 50 \propto (0.025)^{(1-n)} \]

3. **Form Ratios**:
Taking the ratio of the two equations:
\[ \frac{100}{50} = \frac{(0.1)^{(1-n)}}{(0.025)^{(1-n)}} \]
Simplifying the left side gives:
\[ 2 = \frac{(0.1)^{(1-n)}}{(0.025)^{(1-n)}} \]

4. **Expressing the Concentrations**:
We can express \( 0.1 \) and \( 0.025 \) as:
\[ 0.1 = \frac{1}{10} \quad \text{and} \quad 0.025 = \frac{1}{40} \]
Therefore, we can rewrite:
\[ 0.1 = 4 \times 0.025 \]
Thus, we have:
\[ 2 = \left(\frac{0.1}{0.025}\right)^{(1-n)} = (4)^{(1-n)} \]

5. **Solving for \( n \)**:
Taking logarithms or equating the powers gives:
\[ 2 = 4^{(1-n)} \implies 2 = 2^2 \implies 1-n = 2 \implies n = -1 \]

6. **Conclusion**:
The order of the reaction \( n \) is \( \frac{1}{2} \).

### Final Answer:
The correct options are:
1. The order of the reaction is \( \frac{1}{2} \).
2. \( t_{1/2} \) would be \( 100 \sqrt{10} \) minutes for \( [A]_0 = 1 \, M \).