Nuclear Physics And Radioactivity Question 472
Question: A radioactive nucleus A with a half-life T, decays into a nucleus B. At $ t=0, $ there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3. Then, t is given by
Options:
A) $ t=T\log (1.3) $
B) $ t=\frac{T}{\log (1.3)} $
C) $ t=T\frac{\log 2}{\log 1.3} $
D) $ t=\frac{\log 1.3}{\log 2}T $
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Answer:
Correct Answer: D
Solution:
[d] Let initially there are total $ N_{0} $ number of nuclei At time $ t,\frac{N_{B}}{N_{A}}=0.3 $ (given)
$ \Rightarrow ,N_{B}=0.3N_{A} $ $ N_{0}=N_{A}+N_{B}=N_{A}+0.3N_{A}\therefore ,N_{A}=\frac{N_{0}}{1.3} $ As we know $ N_{t}=N_{0}{{e}^{-\lambda t}} $ or, $ \frac{N_{0}}{1.3}=N_{0}{{e}^{-\lambda t}} $ $ \frac{1}{1.3}={{e}^{-\lambda t}}\Rightarrow \ln (1.3)=\lambda t $ or $ t=\frac{\ln (1.3)}{\lambda }\Rightarrow ,t=\frac{\ln (1.3)}{\frac{\ln (2)}{T}}=\frac{\ln (1.3)}{\ln (2)}T $
BETA
