the radioisotope cobalt-60 is used in cancer therapy. the half-life isotope is 5.27 years. which is equation determines the percent of an initial isotope remaining after t years?

Answer : The equation determines the percent of an initial isotope remaining after t years is, [tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]Explanation :Half-life = 5.27 yearsFormula used :[tex]a=\frac{a_o}{2^n}[/tex]        ............(1)where,a = amount of reactant left after n-half lives[tex]a_o[/tex] = Initial amount of the reactantn = number of half livesAnd as we know that,[tex]n=\frac{t}{t_{1/2}}[/tex]        ..........(2)where,t = time[tex]t_{1/2}[/tex] = half-life  = 5.27 yearsNow equating the value of 'n' from (2) to (1), we get:[tex]a=\frac{a_o}{2^{(\frac{t}{t_{1/2}})}}[/tex]       ...........(3)[tex]a=\frac{a_o}{2^{(\frac{t}{5.27})}}[/tex][tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]Therefore, the equation determines the percent of an initial isotope remaining after t years is, [tex]\frac{a}{a_o}\times 100=2^{(-\frac{t}{5.27})}[/tex]