Question

A 46 gram sample of a substance that is used to sterilize surgical instruments has a k-value of 0.1374. Find the substance's half-life in days, round to the nearest tenth.

Answer 1

`t=5.0days`

Explanation:

Using the formula for the exponential decay that is
`N=N_(0)e^(-kt)`, we have N=
`(1)/(2){*}46=23`,
`N_(0)=46` and k=0.1374.

Thus,
`N=N_(0)e^(-kt)` becomes

`23=46{*}e^(-0.1374t)`

`(23)/(46)=e^(-0.1374t)`

`(1)/(2)=e^(-0.1374t)`

Taking log on both sides, we get

`ln((1)/(2))={-0.1374t}`

`t=(ln(1)/(2))/(-0.1)`

`t=(-0.6931)/(-0.1374)`

`t=5.0days`