In an attempt to compare the half-lives of two radioactive elements A and B, a scientist set aside 400 g of each. After 3 months, the scientist found 25 g of A and 200 g of B. Which one of the following statements is true?

Options

A. Half-life of B is twice that of A
B. Half-life of B is four times that of A
C. Half-life of A is twice that of B
D. Half-life of A is four times that of B

Explanation

To solve the problem of comparing the half-lives of two radioactive elements A and B based on the given data, we can follow these steps:

### Step 1: Determine the initial and remaining amounts of A and B
- Initial amount of A (A₀) = 400 g
- Remaining amount of A after 3 months (A) = 25 g
- Initial amount of B (B₀) = 400 g
- Remaining amount of B after 3 months (B) = 200 g

### Step 2: Calculate the decay constant (k) for element A
The formula for the decay constant \( k \) is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A} \right) \]
For element A:
- Time \( t = 3 \) months
- \( A_0 = 400 \) g
- \( A = 25 \) g

Substituting the values:
\[ k_A = \frac{2.303}{3} \log \left( \frac{400}{25} \right) \]
Calculating \( \frac{400}{25} = 16 \):
\[ k_A = \frac{2.303}{3} \log(16) \]
Using \( \log(16) = 4 \log(2) \) (since \( 16 = 2^4 \)):
\[ k_A = \frac{2.303}{3} \times 4 \log(2) \]
Using \( \log(2) \approx 0.301 \):
\[ k_A = \frac{2.303 \times 4 \times 0.301}{3} \approx 0.924 \text{ month}^{-1} \]

### Step 3: Calculate the decay constant (k) for element B
Using the same formula for element B:
\[ k_B = \frac{2.303}{3} \log \left( \frac{400}{200} \right) \]
Calculating \( \frac{400}{200} = 2 \):
\[ k_B = \frac{2.303}{3} \log(2) \]
Substituting \( \log(2) \approx 0.301 \):
\[ k_B = \frac{2.303 \times 0.301}{3} \approx 0.231 \text{ month}^{-1} \]

### Step 4: Calculate the half-lives (T₁/₂) for A and B
The half-life is given by:
\[ T_{1/2} = \frac{0.693}{k} \]
For element A:
\[ T_{1/2, A} = \frac{0.693}{0.924} \approx 0.75 \text{ months} \]
For element B:
\[ T_{1/2, B} = \frac{0.693}{0.231} \approx 3 \text{ months} \]

### Step 5: Compare the half-lives
Now we can compare the half-lives:
\[ \frac{T_{1/2, B}}{T_{1/2, A}} = \frac{3}{0.75} = 4 \]
This means that the half-life of element B is 4 times that of element A.

### Conclusion
The correct statement is that the half-life of element B is 4 times that of element A.