Question

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 1

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.

Answer 2

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))}`