Question

Krypton-91 is a radioactive substance that decays very quickly. The function Q(t)=Qoe^-kt models radioactive decay of krypton-91. Q represents the quantity remaining after t seconds and the decay constant k is approximately 0.07. How long will it take a quantity of krypton-91 to decay t o10% of its origional amount? round your answer to the nearest second.

Answer 1

33 seconds (rounded)

Explanation:

Use the
`Q(t)=Q_(0) e^(-kt)` equation and plug in the given data. You know
`Q_(0)` is 1 and Q has to be 10% of that, or 0.1, and now we just need to find t. Use
`0.1=1e^(-0.07t)` and take the natural log of both sides to get
`ln(0.1)=-0.07t`. After this, we can isolate t set it equal to
`t=(ln(0.1))/(-0.07)`. From here we just simplify and get the decimal answer of t = 32.894, but the question wants us to round to the nearest second, so change that to 33. Hope this helps

Answer 2

32.894 seconds ≈ 33 seconds

Explanation:

Function :
`Q(t)=Q_0 e^(-kt)`

The function models radioactive decay of krypton-91.

The decay constant k is approximately 0.07.

`Q_0` denotes initial amount .

We are require to find How long will it take a quantity of krypton-91 to decay to 10% of its original amount

`\Rightarrow 10\%* Q_0`

`\Rightarrow 0.10* Q_0`

So, we neet to find t at which
`Q(t)=0.10 Q_0`

`0.10 Q_0=Q_0 e^(-0.07t)`

`0.10 =e^(-0.07t)`

Taking log both sides

`\log0.10 =-0.07t \log e`

`\log0.10 = -0.07t * 0.434294481903`

`-1 = -0.0304006137332t`

`(-1)/(-0.0304006137332) = t`

`32.894= t`

Thus it will take 32.894 seconds ≈ 33 seconds a quantity of krypton-91 to decay to 10% of its original amount.