A) 3.45 ´ 1010
B) 1 ´ 1010
C) 3.45 ´ 1015
D) 2.75 ´ 1011
Correct Answer: A
Solution :
Activity of substance that has 2000 disintegration/sec \[=\frac{2000}{3.7\times {{10}^{10}}}=0.054\times {{10}^{-6}}ci=0.054\,\mu ci\] The number of radioactive nuclei having activity A \[N=\frac{A}{\lambda }=\frac{2000\times {{T}_{1/2}}}{{{\log }_{e}}2}\] \[=\frac{2000\times 138.6\times 24\times 3600}{0.693}=3.45\times {{10}^{10}}\]You need to login to perform this action.
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