Question

Suppose that a certain radioactive element decays in such a way that every twenty years the mass of a sample of the element is one third of the initial mass. Given a 100 gram sample of the element, how much of the element remains after 17 years

Answer 1

Answer:

Therefore 21.09 gram of the element remains after 17 years.

Step-by-step explanation:

The decay rate is proportional to the number of nuclei.

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

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

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

Integrating both sides

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

$\Rightarrow ln N= -\lambda t+C$

Initially N=$N_0$ , when t=0

$ln N_0= -\lambda .0+C$

$\Rightarrow C=ln \ N_0$

The equation becomes

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

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

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

$\Rightarrow N=N_0e^{-\lambda t}$

N= Remaining mass after t time

$N_0$= initial mass of the sample

$\lambda$= decay constant

Given that, every 12 years the mass of a sample of a the element is one third of the initial mass.

Here, $N=\frac13N_0$, t= 12 years

$\therefore \frac13N_0=N_0e^{-12\lambda }$

$\Rightarrow e^{-12\lambda }=\frac13$

$\Rightarrow ln e^{-12\lambda }=ln\frac13$                    

$\Rightarrow ln e^{-12\lambda }=ln1-ln3$            [ $ln\frac ab = ln a- ln b$ ]

$\Rightarrow {-12\lambda }= -ln 3$                      [$ln e^a=a$ and $ln 1=0$]

$\Rightarrow\lambda }=\frac{ ln 3}{12}$

The equation becomes

$\therefore N=N_0e^{-\frac{ln3}{12} t}$

Given that, the initial amount $N_0= 100 \ gram$ and t =17

$\therefore N=100e^{-\frac{ln3}{12} 17}$

$\Rightarrow N=21.09$ gram

Therefore 21.09 gram of the element remains after 17 years.