Question

# The activity of a nuclide is 15millicurie. If its decay constant is 0.005 sec−1, the number of atoms present in it is

A
11.1 x 1010
B
1 x 109
C
111 x 1010
D
1.1 x 1010
Solution
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#### Decay constant =0.005sec−1=λActivity of nuclei =15×3.7×107dps[∵1millicurie=3.7×107dps]=5.55×108dps

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