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

Answer:

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

Step-by-step explanation:

Radioactive Decay:

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

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

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

Integrating both sides

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

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

$\Rightarrow N= e^{\lambda t+c_1}$

$\Rightarrow N= e^{\lambda t}.e^{c_1}$

$\Rightarrow N=c e^{\lambda t}$                  [ $e^{c_1}=c$ ]

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

$N=N_0e^{\lambda t}$

$\Rightarrow 70 =140e^{\lambda \times 25}$

$\Rightarrow e^{\lambda \times 25}=\frac{70}{140}$

$\Rightarrow ln|e^{\lambda \times 25}|=ln|\frac{1}{2}|$

$\Rightarrow \lambda \times 25}=-ln|2|$                [ $ln |\frac12|=ln 1-ln 2=0-ln 2=-ln 2$  ]

$\Rightarrow \lambda =-\frac{ln|2|}{25}$

The decay equation becomes

$N=N_0e^{-\frac{ln|2|}{25}t}$

Now putting t= 35

$N=140 e^{-\frac{ln|2|}{25}.35}$

   = 53.05 mg

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