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Q.

The decay constant of Ra226 is 1.37×10−11s−1. A sample  of Ra226 having an activity of 1.5 millicurie will contain

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a

4.05×1018 atoms

b

3.7×1017 atoms

c

2.05×1015 atoms

d

4.7×1010 atoms

answer is A.

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Detailed Solution

1 millicurie =3.7×107disintegrations per second1.5 millicurie = 5.55 × 107 disintegrations per secondor 5.55×107N0=K=1.37×10−11∴N0=5.55×1071.37×10−11=4.05×1018
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