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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?

User Harryghgim
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Final answer:

The equation that determines the percent of an initial isotope remaining after t years is: Percent remaining = (100) x (1/2)^(t/h), where t is the number of years and h is the half-life of the isotope.

Step-by-step explanation:

The equation that determines the percent of an initial isotope remaining after t years is: Percent remaining = (100) x (1/2)(t/h), where t is the number of years and h is the half-life of the isotope.

For example, for the radioisotope cobalt-60 with a half-life of 5.27 years, you can use the equation Percent remaining = (100) x (1/2)(t/5.27) to determine the percent remaining after t years.

Let's say you want to find the percent remaining after 10 years, you would substitute t with 10 in the equation, like this: Percent remaining = (100) x (1/2)(10/5.27). Simplify the equation to find the answer.

User BillT
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Answer : The equation determines the percent of an initial isotope remaining after t years is,
(a)/(a_o)* 100=2^{(-(t)/(5.27))}

Explanation :

Half-life = 5.27 years

Formula used :


a=(a_o)/(2^n) ............(1)

where,

a = amount of reactant left after n-half lives


a_o = Initial amount of the reactant

n = number of half lives

And as we know that,


n=(t)/(t_(1/2)) ..........(2)

where,

t = time


t_(1/2) = half-life = 5.27 years

Now equating the value of 'n' from (2) to (1), we get:


a=\frac{a_o}{2^{((t)/(t_(1/2)))}} ...........(3)


a=\frac{a_o}{2^{((t)/(5.27))}}


(a)/(a_o)* 100=2^{(-(t)/(5.27))}

Therefore, the equation determines the percent of an initial isotope remaining after t years is,
(a)/(a_o)* 100=2^{(-(t)/(5.27))}

User Binil Surendran
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