The decay constant of `Ra^(226)` is `1.37xx10^(-11)s^(-1)`. A sample of `Ra^(226)` having an activity of `1.5` millicurie will contain

Options

A. `4.05xx10^(18)` atoms
B. `3.7xx10^(17)` atoms
C. `2.05xx10^(15)` atoms
D. `4.7 xx 10^(10)` atoms

Answer

A

Explanation

`1 millicurie=3.7xx10^(7)` disintegrations per second
`1.5 millicurie =5.55xx10^(7)` disintegrations per second
or `(5.55xx10^(7))/(N_(0))=K=1.37xx10^(-11)`
`:. N_(0)=(5.55xx10^(7))/(1.37xx10^(-11))=4.05xx10^(18)`