Nuclear Physics And Radioactivity Question 75
Question: The rate of disintegration was observed to be $ 10^{17} $ disintegrations per sec when its half life period is 1445 years. The original number of particles are [Pb. PET 2001]
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:
Rate of disintegration $ \frac{dN}{dt}=10^{17}{{s}^{-1}} $ Half life $ {T_{1/2}}=1445 $ year = 1445 ´ 365 ´ 24 ´ 60 ´60 = 4.55 ´ 1010 sec Now decay constant $ \lambda =\frac{0.693}{{T_{1/2}}}=\frac{0.693}{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} $
Þ N0 = 6.6 ´ 1027.27
BETA
