Nuclear Physics And Radioactivity Question 104
Question: The rate of disintegration was observed to be 1017 disintegrations per sec when its half-life period is 1445 years. The original number of particles are
Options:
A) $ 8.9\times 10^{27} $
B) $ 6.6\times 10^{27} $
C) $ 1.4\times 10^{16} $
D) $ 1.2\times 10^{17} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Rate of disintegration $ \frac{dN}{dt}=10^{17}{{s}^{-1}} $ Half-life $ {T_{1/2}}=1445 $ year $ =1445\times 365\times 24\times 60\times 60=4.55\times 10^{10}\sec $ Now decay constant $ \lambda =\frac{0.693}{{T_{1/2}}}=\frac{0.93}{4.55\times 10^{10}}=1.5\times {{10}^{-11}} $ Per sec The rate of disintegration $ \frac{dN}{dt}=\lambda \times N_{0}\Rightarrow 10^{17}=1.5\times {{10}^{-11}}\times N_{0} $
$ \Rightarrow N_{0}=6.6\times 10^{27} $
BETA
