The half-life of iodine-123 is about 13 hours. You begin with 52 grams of iodine-123. (a) Write an equation that gives the amount of iodine-123, I , remaining after t hours. Write your answer in the form I ( t ) = a ⋅ b t . Round your answer for b to three decimal places.

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hmaritza149 hmaritza149

03-03-2018
Mathematics

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The half-life of iodine-123 is about 13 hours. You begin with 52 grams of iodine-123. (a) Write an equation that gives the amount of iodine-123, I , remaining after t hours. Write your answer in the form I ( t ) = a ⋅ b t . Round your answer for b to three decimal places.

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calculista calculista · Answer 1 · 2018-03-13T15:56:41+01:00

the complete question is
The half-life of iodine-123 is about 13 hours. You begin with 52 grams of iodine-123. (a) Write an equation that gives the amount of iodine-123, I , remaining after t hours. Write your answer in the form I ( t ) = a ⋅ b^ t . (b) what percentage of iodine-123 remains in a sample after 36 hours? Round your answer for b to three decimal places.

Part a)
we know that
exponent equation formula would be an= a0( 1+r)^tand
In this question, the radio isotopes will become half of its weight after the half-life.
So
the ratio would be -0.5 (r= -0.5)The halftime is 13 hours, so the time will be divided by that number
an= a0( 1+r)^t
an= a0( 1-0.5)^(t/13)
an= a0( 1/2)^(t/13)
a0=52 gr
then
an= 52( 1/2)^(t/13)

the answer part A) is an= 52( 1/2)^(t/13)

Part B)
for t=36 hours

an= 52( 1/2)^(36/13)-----------> an= 52( 1/2)^(2.769)=52*(0.147)
this calculation says is that after 36 hours only 0.147 of the initial mass is left

0.147---------> 14.7% of the initial mass (52 gr) is left

the answer part B) is
after 36 hours remains the 14.7% of the initial mass