Respuesta :
Answer:
4492 years (to the nearest year)
Step-by-step explanation:
The quantity of radium left after a time t can be modeled using the equation below:
[tex]Q(t)=Q_0(\frac{1}{2})^{\frac{t}{t_{1/2}} }\\Q_0 =$Initial Amount$\\t_{1/2}=$Half-Life of the Substance$\\t=$Time elapsed$[/tex]
Given that:
[tex]Q_0 =70mg\\t_{1/2}=1600 years[/tex]
The amount of radium left after a t years is:
[tex]Q(t)=70(\frac{1}{2})^{\frac{t}{1600} }[/tex]
If the quantity left is 10 mg, then:
[tex]10=70(\frac{1}{2})^{\frac{t}{1600} }\\\frac{10}{70} =(\frac{1}{2})^{\frac{t}{1600} }\\[/tex]
Changing to logarithm form
[tex]Log_{0.5}\frac{10}{70}=\frac{t}{1600} \\\frac{Log(10/70)}{Log 0.5} =\frac{t}{1600}\\t=\frac{Log(10/70)}{Log 0.5} X 1600\\=4491.8 \approx 4492 \:years\text{ (to the nearest year)}[/tex]
In 4492 years, 10 mg of Radium will be left.
a). Formula for the final amount → [tex]Q(t)=70(0.99956687)^t[/tex]
b). 10 mg will be left after 4492 years.
Exponential decay:
- Exponential decay of a radioactive element is given by the
expression,
[tex]Q(t)=Q_0(1-r)^t[/tex]
Here, [tex]Q(t)=[/tex] Final amount
[tex]Q_0=[/tex] Initial amount
[tex]r=[/tex] Rate of decay
[tex]t=[/tex] Duration of decay
Given in the question,
- Half life of Radium = 1600 years
- Initial amount = 70 mg
a). For [tex]Q(t)=\frac{Q_0}{2}[/tex],
[tex]\frac{Q_0}{2}=Q_0(1-r)^{1600}[/tex]
[tex]0.5=(1-r)^{1600}[/tex]
0.99956687 = 1 - r
r = 0.000433123
r ≈ 0.0433123%
Therefore, exponential function representing the radioactive decay will be,
[tex]Q(t)=70(0.9995668)^t[/tex]
b). For [tex]Q(t)=10\text{ mg}[/tex],
[tex]10=70(0.9995668)^t[/tex]
[tex](0.9995668)^t=\frac{1}{7}[/tex]
[tex]\text{log}(0.9995668)^t=\text{log}(\frac{1}{7})[/tex]
[tex]t.\text{log}(0.9995668)=-\text{log}(7)[/tex]
t ≈ 4492 years
Therefore, a). Formula for the final amount → [tex]Q(t)=70(0.99956687)^t[/tex]
b). 10 mg will be left after 4492 years.
Learn more about the exponential decay here,
https://brainly.com/question/3499464?referrer=searchResults
