Question

Element X decays radioactively with a half life of 11 minutes. If there are 870 grams of Element X, how long, to the nearest tenth of a minute, would it take the element to decay to 154 grams?

Answer 1

Answer:

It would take 27.5 minutes the element to decay to 154 grams.

Step-by-step explanation:

The decay equation:

$\frac {dN}{dt}\propto -N$

$\Rightarrow \R\frac {dN}{dt}=-\lambda N$

$\Rightarrow \frac {dN}N=-\lambda dt$

Integrating both sides

$\Rightarrow \int \frac {dN}N=\int-\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

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

$ln|N_0|=-\lambda .0+c$

$\Rightarrow c=ln|N_0|$

$ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$

Decay equation:              

                    $ln|\frac{N}{N_0}|=-\lambda t$

Given that, the half life of of element X is 11 minutes.

For half life, $N=\frac12 N_0$,  t= 11 min.

$ln|\frac{N}{N_0}|=-\lambda t$

$\Rightarrow ln|\frac{\frac12N_0}{N_0}|=-\lambda . 11$

$\Rightarrow ln|\frac12}|=-\lambda . 11$

$\Rightarrow -\lambda . 11=ln|\frac12}|$

$\Rightarrow \lambda =\frac{ln|\frac12|}{-11}$

$\Rightarrow \lambda =\frac{ln|2|}{11}$                [ $ln|\frac12|=ln|1|-ln|2|=-ln|2|$ , since ln|1|=0]

N=154 grams, $N_0$ = 870 grams, t=?

$ln|\frac{N}{N_0}|=-\lambda t$

$\Rightarrow ln|\frac{154}{870}|=-\frac{ln|2|}{11}.t$

$\Rightarrow t= \frac{ln|\frac{154}{870}|\times 11}{-ln|2|}$

      =27.5 minutes

It would take 27.5 minutes the element to decay to 154 grams.