A

To find the average velocity during the specified time interval for a particle whose velocity varies according to the equation \( v = v_0 e^{-\lambda t} \), we proceed with the following steps:

1. Determine the time intervals: The velocity decreases from \( v_0 \) to \( \frac{v_0}{2} \). Setting up the equation:
2. \( \frac{v_0}{2} = v_0 e^{-\lambda t} \)
3. Dividing by \( v_0 \) gives:
4. \( \frac{1}{2} = e^{-\lambda t} \)
5. Taking the natural logarithm on both sides:
6. \( -\lambda t = \ln(\frac{1}{2}) \)
7. Thus, we find \( t = -\frac{\ln(1/2)}{\lambda} = \frac{\ln(2)}{\lambda} \).
8. Integrate to find displacement: The displacement \( s \) between the time intervals \( 0 \) to \( t \) is given by integrating the velocity:
9. \( s = \int_0^t v dt = \int_0^t v_0 e^{-\lambda t} dt. \)
10. Evaluating this integral:
11. \( s = v_0 \left[-\frac{1}{\lambda} e^{-\lambda t} \right]_0^t = \frac{v_0}{\lambda} (1 - e^{-\lambda t}) \).
12. Substituting \( t = \frac{\ln(2)}{\lambda} \) gives:
13. \( s = \frac{v_0}{\lambda} (1 - e^{-\ln(2)}) = \frac{v_0}{\lambda} (1 - \frac{1}{2}) = \frac{v_0}{2\lambda} \).

Average velocity: Average velocity \( v_{avg} \) is given by displacement over time:
\( v_{avg} = \frac{s}{t} = \frac{\frac{v_0}{2\lambda}}{\frac{\ln(2)}{\lambda}} = \frac{v_0}{2 \ln(2)}. \)
Therefore, average velocity during the time interval is \( \frac{v_0}{2 \ln(2)}. \) Thus, the final answer matches with option A (if given).