Nuclear Physics And Radioactivity Question 312
Question: Half lives of two radioactive substances A and B are respectively 20 minutes and 40 minutes. Initially the sample of A and B have equal number of nuclei. After 80 minutes, the ratio of remaining number of A and B nuclei is [CBSE PMT 1998; JIPMER 2000]
Options:
A) 1 : 16
B) 4 : 1
C) 1 : 4
D) 1 : 1
Show Answer
Answer:
Correct Answer: C
Solution:
For 80 minutes, number of half-lives of sample $ A=n_{A}=\frac{80}{20}=4 $ and number of half-lives of sample $ B=n_{B}=\frac{80}{40}=2. $ Also by using $ N=N_{0}{{( \frac{1}{2} )}^{n}} $
Þ $ N\propto \frac{1}{2^{n}} $
Þ $ \frac{N_{A}}{N_{B}}=\frac{{{2}^{n_{B}}}}{{{2}^{n_{A}}}}=\frac{2^{2}}{2^{4}}=\frac{1}{4} $
BETA
