Respuesta :
Answer:
Therefore 53.05 mg will remain of the given radioactive substance after 35 hours.
Step-by-step explanation:
Radioactive Decay:
[tex]\frac{dN}{dt}\propto N[/tex]
[tex]\Rightarrow \frac{dN}{dt}=\lambda N[/tex]
[tex]\Rightarrow \frac{dN}{N}=\lambda dt[/tex]
Integrating both sides
[tex]\int \frac{dN}{N}=\int\lambda dt[/tex]
[tex]\Rightarrow ln |N|= \lambda t+c_1[/tex]
[tex]\Rightarrow N= e^{\lambda t+c_1}[/tex]
[tex]\Rightarrow N= e^{\lambda t}.e^{c_1}[/tex]
[tex]\Rightarrow N=c e^{\lambda t}[/tex] [ [tex]e^{c_1}=c[/tex] ]
When t=0, [tex]N=N_0[/tex]= initial amount
[tex]N_0=c e^{\lambda .0}[/tex]
[tex]\Rightarrow N_0=c[/tex]
Therefore the decay equation is
[tex]N=N_0e^{\lambda t}[/tex]
Given that, [tex]N_0[/tex]= Initial amount of the radioactive substance= 140 mg
After 25 hours, 70 mg of substance remains.
N= 70 mg, t=25 hours
[tex]N=N_0e^{\lambda t}[/tex]
[tex]\Rightarrow 70 =140e^{\lambda \times 25}[/tex]
[tex]\Rightarrow e^{\lambda \times 25}=\frac{70}{140}[/tex]
[tex]\Rightarrow ln|e^{\lambda \times 25}|=ln|\frac{1}{2}|[/tex]
[tex]\Rightarrow \lambda \times 25}=-ln|2|[/tex] [ [tex]ln |\frac12|=ln 1-ln 2=0-ln 2=-ln 2[/tex] ]
[tex]\Rightarrow \lambda =-\frac{ln|2|}{25}[/tex]
The decay equation becomes
[tex]N=N_0e^{-\frac{ln|2|}{25}t}[/tex]
Now putting t= 35
[tex]N=140 e^{-\frac{ln|2|}{25}.35}[/tex]
= 53.05 mg
Therefore 53.05 mg will remain of the given radioactive substance after 35 hours.
