Question

A radioactive substance decays exponentially. A scientist begins with 140 milligrams of a radioactive substance. After 25 hours, 70 mg of the substance remains. How many milligrams will remain after 35 hours

Answer 1

Therefore 53.05 mg will remain of the given radioactive substance after 35 hours.

Explanation:

Radioactive Decay:

`(dN)/(dt)\propto N`

`\Rightarrow (dN)/(dt)=\lambda N`

`\Rightarrow (dN)/(N)=\lambda dt`

Integrating both sides

`\int (dN)/(N)=\int\lambda dt`

`\Rightarrow ln |N|= \lambda t+c_1`

`\Rightarrow N= e^(\lambda t+c_1)`

`\Rightarrow N= e^(\lambda t).e^(c_1)`

`\Rightarrow N=c e^(\lambda t)` [
`e^(c_1)=c` ]

When t=0,
`N=N_0`= initial amount

`N_0=c e^(\lambda .0)`

`\Rightarrow N_0=c`

Therefore the decay equation is

`N=N_0e^(\lambda t)`

Given that,
`N_0`= Initial amount of the radioactive substance= 140 mg

After 25 hours, 70 mg of substance remains.

N= 70 mg, t=25 hours

`N=N_0e^(\lambda t)`

`\Rightarrow 70 =140e^(\lambda * 25)`

`\Rightarrow e^(\lambda * 25)=(70)/(140)`

`\Rightarrow ln|e^(\lambda * 25)|=ln|(1)/(2)|`

`\Rightarrow \lambda * 25}=-ln|2|` [
`ln |\frac12|=ln 1-ln 2=0-ln 2=-ln 2` ]

`\Rightarrow \lambda =-(ln|2|)/(25)`

The decay equation becomes

`N=N_0e^)/(25)t`

Now putting t= 35

`N=140 e^-(ln`

= 53.05 mg

Therefore 53.05 mg will remain of the given radioactive substance after 35 hours.