Respuesta :
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 years
Formula 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 reactant
n = number of half lives
And 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 years
Now 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]
