Nuclear Physics And Radioactivity Question 72
Question: Half life of a radio-active substance is 20 minutes. The time between 20% and 80% decay will be [KCET 2003]
Options:
A) 20 minutes
B) 40 minutes
C) 30 minutes
D) 25 minutes
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ {T_{1/2}}=20 $ minutes; we know $ \frac{N}{N_{0}}={{( \frac{1}{2} )}^{t/{T_{1/2}}}} $ For 20% decay $ \frac{N}{N_{0}}=\frac{80}{100}={{( \frac{1}{2} )}^{t_{1}/20}} $ ….. (i) For 80% decay $ \frac{N}{N_{0}}=\frac{20}{100}={{( \frac{1}{2} )}^{t_{2}/20}} $ ….. (ii) Dividing (ii) by (i) $ \frac{1}{4}={{( \frac{1}{2} )}^{\frac{(t_{2}-t_{1})}{20}}}; $ on solving we get $ t_{2}-t_{1}=40 $ min.
BETA
