A radioactive sample decays by `63%` of its initial value in `10s`. It would have decayed by `50%` of its initial value in .

Options

A. `7s`
B. `14 s`
C. `0.7s`
D. `1.4 s`

Explanation

To solve the problem of how long it takes for a radioactive sample to decay by 50% of its initial value, given that it decays by 63% in 10 seconds, we can follow these steps:

### Step 1: Understand the decay percentage
We know that the sample decays by 63% in 10 seconds. This means that 37% of the sample remains after 10 seconds.

### Step 2: Use the decay formula
The decay of a radioactive sample can be described by the equation:
\[ N = N_0 e^{-\lambda t} \]
where:
- \( N \) is the remaining quantity of the sample.
- \( N_0 \) is the initial quantity of the sample.
- \( \lambda \) is the decay constant.
- \( t \) is the time.

### Step 3: Set up the equation for the given decay
After 10 seconds, we can express the remaining amount as:
\[ N = N_0 \times 0.37 \]
Substituting into the decay formula:
\[ N_0 \times 0.37 = N_0 e^{-\lambda \cdot 10} \]
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ 0.37 = e^{-\lambda \cdot 10} \]

### Step 4: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \ln(0.37) = -\lambda \cdot 10 \]
Calculating \( \ln(0.37) \):
\[ \ln(0.37) \approx -1 \]

### Step 5: Solve for \( \lambda \)
From the equation:
\[ -1 = -\lambda \cdot 10 \]
Thus:
\[ \lambda = \frac{1}{10} \, \text{s}^{-1} \]

### Step 6: Calculate the half-life
The half-life \( t_{1/2} \) can be calculated using the formula:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting \( \lambda \):
\[ t_{1/2} = \frac{\ln(2)}{\frac{1}{10}} \]
Calculating \( \ln(2) \):
\[ \ln(2) \approx 0.693 \]
So:
\[ t_{1/2} = 0.693 \times 10 \]
\[ t_{1/2} \approx 6.93 \, \text{s} \]

### Step 7: Conclusion
Therefore, the time it takes for the sample to decay by 50% of its initial value is approximately **7 seconds**.

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