Mean life of a radioactive sample is `t_0`. What fraction of sample remains left after time `t_0ln_2`?

Options

A. `1/4`
B. `1/2`
C. `3/5`
D. `2/3`

Explanation

To solve the problem of finding the fraction of a radioactive sample that remains after a time of \( t_0 \ln 2 \), where \( t_0 \) is the mean life of the sample, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Mean Life and Decay Constant Relationship**:
The mean life \( t_0 \) of a radioactive sample is related to the decay constant \( \lambda \) by the formula:
\[ t_0 = \frac{1}{\lambda} \]
Therefore, we can express the decay constant as:
\[ \lambda = \frac{1}{t_0} \]

2. **Relate Time to the Number of Nuclei**:
The number of radioactive nuclei \( N(t) \) remaining after time \( t \) can be expressed using the exponential decay formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei.

3. **Substitute for \( t \)**:
We need to find the fraction remaining after a time of \( t = t_0 \ln 2 \). Substituting \( t \) and \( \lambda \) into the equation gives:
\[ N(t_0 \ln 2) = N_0 e^{-\left(\frac{1}{t_0}\right)(t_0 \ln 2)} \]

4. **Simplify the Exponential**:
Simplifying the exponent:
\[ N(t_0 \ln 2) = N_0 e^{-\ln 2} \]
Using the property of exponents, \( e^{-\ln 2} = \frac{1}{2} \):
\[ N(t_0 \ln 2) = N_0 \cdot \frac{1}{2} \]

5. **Calculate the Fraction Remaining**:
The fraction of the sample that remains is given by:
\[ \text{Fraction remaining} = \frac{N(t_0 \ln 2)}{N_0} = \frac{N_0 \cdot \frac{1}{2}}{N_0} = \frac{1}{2} \]

### Final Answer:
The fraction of the radioactive sample that remains after time \( t_0 \ln 2 \) is:
\[ \frac{1}{2} \]