A radiation level of 5 rems per year is considered to be safe for humans. A part

Question

A radiation level of 5 rems per year is considered to be safe for humans. A particular site is polluted by a radioactive substance at a level of 100 rems per year. The radioactive material has a half-life of 10 years. (a) [2 points] Determine the exact value of the decay constant k of the radioactive material. Notation: Let Q(t) be the level of radiation (in rems per year) at t years into the future. So Q0 = Q(0) is the current level of radiation. (b) [1 point] Write out an exact formula for the level of radiation Q(t) (in rems per year) that will be present at the site after t years. (Do not use a calculator for parts (a) and (b) of this question). (c) [2 points] About how many years will the site have to be sealed off before it is safe for humans to use it again? You may use a calculator here!
A radiation level of 5 rems per year is considered to be safe for humans. A particular site is polluted by a radioactive substance at a level of 100 rems per year. The radioactive material has a half-life of 10 years (a) [2 points] Determine the exact value of the decay constant k of the radioactive material. Notation: Let Q(t) be the level of radiation (in rems per year) at t years into the future. So Qo -9(0) is the current level of radiation. (b) [1 point] Write out an exact formula for the level of radiation Q(t) (in rems per year) that will be present at the site after t years. (Do not use a caleulator for parts (a) and (b) of this question). (e) [2 pointa] About how many years will the site have to be sealed off before it is safe for humans to use it again? You may use a calculator here!

Explanation / Answer

(a)k = ln2 / t1/2 = ln2 /10 = 0.0693

(b)

Since it is a first order reaction

Q(t) = Q(0) e-kt

=> Q(t) = 100e-0.0693t

(c)

Q(t) = 5 , we have to find t

Q(t) = 100e-0.0693t

=> 5 = 100 e-0.0693t

=> 0.05 = e-0.0693t

=> ln(0.05) = -0.0693t

=> t = ln(0.05) / -0.0693 = 43.2285 years

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