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If one starts with 1 Curie (Ci) of radio...

If one starts with `1` Curie `(Ci)` of radioactive substance `(t_(1//2)=15 hr)` the activity left after a periof of two weeks will be about `0.02x muCi`. Find the value of `x`.

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The correct Answer is:
9
`k=0.693/(15 hr)=0.0462 hr^(-1)`
`k=2.3/(14xx24 hr) log c_(0)/c_(t)`
`0.0462 hr^(-1)=2.3/(14xx24 hr) log.(1 Ci)/c_(t)`
Solve for `c_(t):`
`:. c_(t)=1.82xx10^(-7) Ci~~0.18 muCi=0.02x muCi`
`:. x=9`
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