Respuesta :
Answer:
We can use the sample about 42 days.
Step-by-step explanation:
Decay Equation:
[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[/tex]
When t=0, N=[tex]N_0[/tex] = initial amount
[tex]\Rightarrow ln|N_0|=-\lambda .0+c[/tex]
[tex]\Rightarrow c= ln|N_0|[/tex]
[tex]\therefore 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].......(1)
[tex]\frac{N}{N_0}=e^{-\lambda t}[/tex].........(2)
Logarithm:
- [tex]ln|\frac mn|= ln|m|-ln|n|[/tex]
- [tex]ln|ab|=ln|a|+ln|b|[/tex]
- [tex]ln|e^a|=a[/tex]
- [tex]ln|a|=b \Rightarrow a=e^b[/tex]
- [tex]ln|1|=0[/tex]
130 days is the half-life of the given radioactive element.
For half life,
[tex]N=\frac12 N_0[/tex], [tex]t=t_\frac12=130[/tex] days.
we plug all values in equation (1)
[tex]ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130[/tex]
[tex]\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130[/tex]
[tex]\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130[/tex]
[tex]\rightarrow -ln|2|=-\lambda \times 130[/tex]
[tex]\rightarrow \lambda= \frac{-ln|2|}{-130}[/tex]
[tex]\rightarrow \lambda= \frac{ln|2|}{130}[/tex]
We need to find the time when the sample remains 80% of its original.
[tex]N=\frac{80}{100}N_0[/tex]
[tex]\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t[/tex]
[tex]\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t[/tex]
[tex]\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t[/tex]
[tex]\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}[/tex]
[tex]\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}[/tex]
[tex]\Rightarrow t\approx 42[/tex]
We can use the sample about 42 days.
