Nuclear Physics And Radioactivity Question 69
Question: Half-life of a radioactive substance is 20 minutes. Difference between points of time when it is 33% disintegrated and 67% disintegrated is approximately [AIIMS 2000]
Options:
A) 10 min
B) 20 min
C) 30 min
D) 40 min
Show Answer
Answer:
Correct Answer: B
Solution:
$ \lambda =\frac{0.693}{{T_{1/2}}}=\frac{0.693}{20}=0.03465 $ Now time of decay $ t=\frac{2.303}{\lambda }\log \frac{N_{0}}{N} $
$ \Rightarrow t_{1}=\frac{2.303}{0.03465}\log \frac{100}{67}=11.6\min $ min and $ t_{2}=\frac{2.303}{0.03465}\log \frac{100}{33}=32min $ Thus time difference between points of time = t1 ? t2 =32 ? 11.6 = 20.4 min $ \approx $ 20 min.
BETA
