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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`. |
Answer» Correct Answer - 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`Correct Answer - 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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