S Quiz submissions Quiz x B) Mathematics l Michigan x C CoB Mi Chegg Study I Gui

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S Quiz submissions Quiz x B) Mathematics l Michigan x C CoB Mi Chegg Study I Guided S X Secure h ps://ww2.math msu.edu /webwork2/mt 24 ss17 76260/ Hw03 effectiveUse perry tr18ukey ba659475de81b19ca34b2ea3b4e2a 72 Hw03: Problem 11 Hw03 Problem 11 Previous Problem List Next User Settings Grades (1 poin Get help entering answe Problems One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at an annual rate of approximately 2.4% per year. (a) Write the percent of strontium-90 remaining. P. as a nunction of years. t. since the nuclear accident. Problem 1 Problem 2 Answer: P- e (0.024 00 Problem 3 Problem 4 (b) Estimate the half-life of strontium-90. Problem 5 Problem 6 V Answer: t 22 years. Problem Problem 8 (c) After the Chernobyl disaster, it was predicted that the region would not be safe for human habitation for 95 years. Estimate the percent of Problem 9 original strontium-90 remaining at this time. Problem 10 Answer: 10.23 Problem 1 Problem Note: In the second and third answer blank, round your answers to two decimal places. Problem Note: You can earn partial credit on this problem. Preview My Answers Submit Answers Show me another You have attempted this problem 7 times. Your overa recorded score is 0%. O M 1:48 PM /29/2017

Explanation / Answer

Let the initial amount of strontium be = Po

the halflife equation is

P = Po*e^(-k*t)

where k is the decay constant = .024

as r (decay rate) = 2.4% per year = .024

and decay rate is directly proportional to decay constant

P is the amount of strontium left after t years

=>a> P = Po*e^(- .024t)

b> the half life of an element is the time taken for the element to decay 50% of it initial value

=> P =Po/2

=> Po/2 = Po*e^(-.024t)

1/2 = e^(-.024t)

take log both sides

=> ln(.5) = -.024t

or t = - ln(.5)/.024

hence t = half life = 28.88 years

c> P = Po*e^(-.024*95)

P = 0.10228Po

so after 95 years P is .10228 times of Po

=> percentage of the original strotium remaining at this time is = 10.228 %

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