Answer:
1) 0.25 kg
2) 0.000977 kg
3) 0.707 kg
[tex]\textsf{4)}\quad N(t)=\left(\dfrac{1}{2}\right)^{\dfrac{t}{1600}}[/tex]
Step-by-step explanation:
To determine how much radium will be present after t years, given it has a half-life of about 1600 years, we can use the half-life formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Half Life Formula}}\\\\N(t)=N_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{t_{\frac12}}}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$N(t)$ is the quantity remaining.}\\\phantom{ww}\bullet\;\textsf{$N_0$ is the initial quantity.}\\ \phantom{ww}\bullet\;\textsf{$t$ is the time elapsed.}\\\phantom{ww}\bullet\;\textsf{$t_{\frac12}$ is the half-life of the substance.}\end{array}}[/tex]
In this case:
- [tex]t_{\frac{1}{2}}=1600\; \sf years[/tex]
Substitute the values into the half-life formula to create an equation for the amount of radium present N(t) in kilograms after t years:
[tex]N(t)=1\cdot \left(\dfrac{1}{2}\right)^{\dfrac{t}{1600}}\\\\\\N(t)=\left(\dfrac{1}{2}\right)^{\dfrac{t}{1600}}[/tex]
To calculate how much radium is present after the given number of years, substitute the given values of t into the equation.
Question 1
When t = 3200 years:
[tex]N(3200)=\left(\dfrac{1}{2}\right)^{\dfrac{3200}{1600}}\\\\\\N(3200)=\left(\dfrac{1}{2}\right)^{2}\\\\\\N(3200)=0.25\; \sf kg[/tex]
Question 2
When t = 16000 years:
[tex]N(16000)=\left(\dfrac{1}{2}\right)^{\dfrac{16000}{1600}}\\\\\\N(16000)=\left(\dfrac{1}{2}\right)^{10}\\\\\\N(16000)=0.000977\; \sf kg\;(3\;s.f.)[/tex]
Question 3
When t = 800 years:
[tex]N(16000)=\left(\dfrac{1}{2}\right)^{\dfrac{800}{1600}}\\\\\\N(16000)=\left(\dfrac{1}{2}\right)^{\frac{1}{2}}\\\\\\N(16000)=0.707\; \sf kg\;(3\;s.f.)[/tex]
Question 4
When t = t years:
[tex]N(t)=\left(\dfrac{1}{2}\right)^{\dfrac{t}{1600}}[/tex]
