Chemistry · Chemical Kinetics
JEE Main 2026 — 24 January, Evening Shift — Question 68
The half-life of is days. After days, of original activity remained. The value of in days is . (Nearest integer) (Given: , )
Answer: 102
Numerical answer — enter this value.
Step-by-step solution
Radioactive decay, we have
= \left(\frac{1}{2}\right)^{t/T_{1/2}}$$ $$0.75 = \left(\frac{1}{2}\right)^{x/245}$$ $$\log(0.75) = \frac{x}{245}\log\left(\frac{1}{2}\right)$$ $$\log\left(\frac{3}{4}\right) = 0.4771 - 0.6020 = -0.1249$$ $$\log\left(\frac{1}{2}\right) = -0.3010$$ $$\frac{x}{245} = \frac{-0.1249}{-0.3010} = 0.4147$$x = 101.6 \approx 102
Answer key and solution verified before publishing.
Practise Chemical Kinetics
Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.
- Exam
- JEE Main 2026
- Subject
- Chemistry
- Chapter
- Chemical Kinetics
- Topic
- Integrated Rate Laws