Respuesta :
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.
[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]\Rightarrow \int\frac{dN}{N}=\int-\lambda \ dt[/tex]
[tex]\Rightarrow ln N= -\lambda t+C[/tex]
Initially N=[tex]N_0[/tex] , when t=0
[tex]ln N_0= -\lambda .0+C[/tex]
[tex]\Rightarrow C=ln \ N_0[/tex]
The equation becomes
[tex]ln N=-\lambda t+ln N_0[/tex]
[tex]\Rightarrow ln N-ln N_0=-\lambda t[/tex]
[tex]\Rightarrow \ln\frac{N}{N_0}=-\lambda t[/tex]
[tex]\Rightarrow N=N_0e^{-\lambda t}[/tex]
N= Remaining mass after t time
[tex]N_0[/tex]= initial mass of the sample
[tex]\lambda[/tex]= decay constant
Given that, every 12 years the mass of a sample of a the element is one third of the initial mass.
Here, [tex]N=\frac13N_0[/tex], t= 12 years
[tex]\therefore \frac13N_0=N_0e^{-12\lambda }[/tex]
[tex]\Rightarrow e^{-12\lambda }=\frac13[/tex]
[tex]\Rightarrow ln e^{-12\lambda }=ln\frac13[/tex]
[tex]\Rightarrow ln e^{-12\lambda }=ln1-ln3[/tex] [ [tex]ln\frac ab = ln a- ln b[/tex] ]
[tex]\Rightarrow {-12\lambda }= -ln 3[/tex] [[tex]ln e^a=a[/tex] and [tex]ln 1=0[/tex]]
[tex]\Rightarrow\lambda }=\frac{ ln 3}{12}[/tex]
The equation becomes
[tex]\therefore N=N_0e^{-\frac{ln3}{12} t}[/tex]
Given that, the initial amount [tex]N_0= 100 \ gram[/tex] and t =17
[tex]\therefore N=100e^{-\frac{ln3}{12} 17}[/tex]
[tex]\Rightarrow N=21.09[/tex] gram
Therefore 21.09 gram of the element remains after 17 years.
