Question

The activity of a nuclide is 15millicurie. If its decay constant is 0.005 sec1, 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 =0.005sec1=λ
Activity of nuclei =15×3.7×107dps[1millicurie=3.7×107dps]
=5.55×108dps

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