Answer:
167.27 mg.
Step-by-step explanation:
We have been given that the half-life of Radium-226 is 1590 years and a sample contains 400 mg.
We will use half life formula to solve our given problem.
[tex]N(t)=N_0*(\frac{1}{2})^{\frac{t}{t/2}[/tex], where N(t)= Final amount after t years, [tex]N_0[/tex]= Original amount, t/2= half life in years.
Now let us substitute our given values in half-life formula.
[tex]N(2000)=400*(\frac{1}{2})^{\frac{2000}{1590}[/tex]
[tex]N(2000)=400*(0.5)^{1.2578616352201258}[/tex]
[tex]N(2000)=400*0.4181633028874878239[/tex]
[tex]N(2000)=167.26532115499512956\approx 167.27[/tex]
Therefore, the remaining amount of Radium-226 after 2000 years will be 167.27 mg.
