Answer:
[tex]\textsf{A)}\quad 1.6 \;\sf kg[/tex]
[tex]\textsf{B)}\quad 0.8 \;\sf kg[/tex]
[tex]\textsf{C)}\quad 0.1 \;\sf kg[/tex]
[tex]\textsf{D)}\quad N(t)=3.2 \left(\dfrac{1}{2}\right)^{\dfrac{t}{4}}[/tex]
Step-by-step explanation:
To determine how of a radioactive isotope will be present after t days, given it has a half-life of 4 days, 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]N_0 = 3.2\;\sf kg[/tex]
- [tex]t_{\frac{1}{2}}=4\; \sf days[/tex]
Substitute the values into the half-life formula to create an equation for the amount of radioactive isotope present N(t) in kilograms after t days:
[tex]N(t)=3.2\cdot \left(\dfrac{1}{2}\right)^{\dfrac{t}{4}}[/tex]
To calculate how much radium is present after the given number of days, substitute the given values of t into the equation.
Part A
When t = 4 days:
[tex]N(4)=3.2\cdot \left(\dfrac{1}{2}\right)^{\dfrac{4}{4}}\\\\\\N(4)=3.2\cdot \left(\dfrac{1}{2}\right)^{1}\\\\\\N(4)=3.2\cdot \left(\dfrac{1}{2}\right)\\\\\\N(4)=1.6\; \sf kg[/tex]
Part B
When t = 8 days:
[tex]N(8)=3.2\cdot \left(\dfrac{1}{2}\right)^{\dfrac{8}{4}}\\\\\\N(8)=3.2\cdot \left(\dfrac{1}{2}\right)^{2}\\\\\\N(8)=3.2\cdot \left(\dfrac{1}{4}\right)\\\\\\N(8)=0.8\; \sf kg[/tex]
Part C
When t = 20 days:
[tex]N(20)=3.2\cdot \left(\dfrac{1}{2}\right)^{\dfrac{20}{4}}\\\\\\N(20)=3.2\cdot \left(\dfrac{1}{2}\right)^{5}\\\\\\N(20)=3.2\cdot \left(\dfrac{1}{32}\right)\\\\\\N(20)=0.1\; \sf kg[/tex]
Question D
When t = t years:
[tex]N(t)=3.2 \left(\dfrac{1}{2}\right)^{\dfrac{t}{4}}[/tex]
