If time t is required for a radioactive substance to become one third of its initial amount, what fraction would be left after 0.5 t ?

Options

A. `(1)/(2)`
B. `(1)/(sqrt(3))`
C. `(1)/(3)`
D. `sqrt((2)/(3))`

Answer

To solve the problem step by step, we will use the principles of radioactive decay and the integrated rate law.

### Step-by-Step Solution:

1. **Understanding the Problem**: We know that it takes time \( t \) for a radioactive substance to decay to one third of its initial amount. We need to find out what fraction of the initial amount remains after \( 0.5t \).

2. **Initial Setup**: Let the initial amount of the substance be \( N_0 \). After time \( t \), the amount remaining is given as:
\[ N_t = \frac{N_0}{3} \]

3. **Using the Integrated Rate Law**: The integrated rate law for radioactive decay can be expressed as:
\[ \ln \left( \frac{N_0}{N_t} \right) = kt \]
where \( k \) is the decay constant and \( N_t \) is the amount remaining after time \( t \).

4. **Finding the Decay Constant**: From the previous equation, we can express the decay constant \( k \) for the time \( t \):
\[ \ln \left( \frac{N_0}{N_t} \right) = \ln \left( \frac{N_0}{\frac{N_0}{3}} \right) = \ln(3) \]
Therefore, we have:
\[ kt = \ln(3) \]

5. **Finding the Amount After \( 0.5t \)**: We want to find the amount remaining after \( 0.5t \). We can express this as:
\[ \ln \left( \frac{N_0}{N_{0.5t}} \right) = k(0.5t) \]
Substituting \( kt \) from the previous step, we get:
\[ \ln \left( \frac{N_0}{N_{0.5t}} \right) = 0.5 \ln(3) \]

6. **Simplifying the Equation**: We can rewrite \( 0.5 \ln(3) \) as:
\[ \ln \left( \frac{N_0}{N_{0.5t}} \right) = \ln(\sqrt{3}) \]

7. **Solving for \( N_{0.5t} \)**: By exponentiating both sides, we find:
\[ \frac{N_0}{N_{0.5t}} = \sqrt{3} \]
Rearranging gives:
\[ N_{0.5t} = \frac{N_0}{\sqrt{3}} \]

8. **Finding the Fraction Remaining**: The fraction of the initial amount remaining after \( 0.5t \) is:
\[ \text{Fraction remaining} = \frac{N_{0.5t}}{N_0} = \frac{1}{\sqrt{3}} \]

### Final Answer:
The fraction of the radioactive substance remaining after \( 0.5t \) is \( \frac{1}{\sqrt{3}} \).