Discussion
A quantity decaying with \( dN/dt = -\lambda N \) falls to half its initial value after a time
Answer: C
The solution is \( N = N_{0}e^{-\lambda t} \). Setting \( N = N_{0}/2 \),
\[ e^{-\lambda t} \;=\; \tfrac{1}{2} \quad \Longrightarrow \quad \lambda t \;=\; \ln 2 \quad \Longrightarrow \quad t \;=\; \frac{\ln 2}{\lambda} \;\approx\; \frac{0.693}{\lambda}. \]
The time \( 1/\lambda \) is the mean lifetime, at which the quantity has fallen to \( 1/e \) rather than a half
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development