Question

Iodine 131 is a radioactive isotope that decays as time passes. The function M(t) = 200e^-0.093t gives the mass of Iodine 131 remaining in the sample measured in grams in terms of t number of days since the samples mass was originally measured.

a) What was the initial mass of Iodine 131?

b) What is the one days decay factor and what is the one day percent change?

c) How long will it take for the sample to decay such that only half of the original mass of Iodine 131 still remains?

d) How long will it take for only 35 grams of Iodine 131 to still be present in the sample?

e) What is the mass of the remaining Iodine 131 after one week has passed since the mass was originally measured?

f) What is the one week growth factor, and what is the one week percent change?

g) Write a function B that closely approximates the mass of the remaining Iodine 131 in the samplemeasured in grams in terms of the number of weeks w since the samples mass was originally measured?

h) What is the one hour growth factor, and what is the one hour percent change?

g) Write a function C that closely approximates the mass of the remaining Iodine 131 in the sample measured in grams in terms of the number of hours h since the samples mass was originally measured?

Answer 1

a). When t = 0, we have M(0) = 200 so the mass of Iodine 131 was 200 grams.

b). When t = 0, M(1) = 200e^-0.093. Hence, one day’s decay factor is 0.093. Also, M(1) = 200e^-0.093 = 200*0.9111935 = 182.239 grams ( on rounding off to the nearest milligram). Thus, the one day change is [(200-182.239)/200]*100 % = 8.88 % ( on rounding off to 2 decimal places).

c). Let the sample take t days to decay such that only half of the original mass of Iodine 131 still remains. Then 100 = 200e^-0.093t or, e^-0.093t = 100/200 = ½. Hence -0.093t = ln(1/2) =           -0.069314718 so that t = 0.069314718/0.093 = 7.45319549 , say 7 days (on rounding off to the nearest whole number).

d). Let the sample take t days to decay such that only 35 grams of Iodine 131 is still present in the sample. Then, 35 = 200e^-0.093t or, e^-0.093t = 35/200 = 0.175. Hence -0.093t = ln(0.175) = -1.742969305 so that t = 1.742969305/0.093 = 18.74160543 , say 19 days(on rounding off to the nearest whole number).

e). When t = 7, we have M(7) = 200e^-0.093*7 = 200e^-0.651 = 200* 0.521523991 = 104.305( on rounding off to the nearest milligram). Thus, the mass of the remaining Iodine 131 after one week has passed since the mass was originally measured will be 104.305 grams.

f). When t = 7, M(7) = 200e^-0.093*7 = 200e^-0.651. Hence, one week’s growth factor is -0.651. Also, M(7) =200e^-0.651 =200* 0.521523991 = 104.305 grams ( on rounding off to the nearest milligram). Thus, the one week’s change is [(200-104.305)/200]*100 % =47.85 %( on rounding off to 2 decimal places).

g). Since 1 week = 7 days, w weeks = 7t days so that t would be replaced by 7w . Hence, B = 200e^-0.093*7w = 200e^-0.651w , where B is the mass of the remaining Iodine 131 in the sample measured in grams in terms of the number of weeks w since the samples mass was originally measured.

h). When t = 1/24, we have M(1h) = 200e^-0.093*1/24 = 200e^-0.003875 . Hence, one hour’s growth factor is – 0.003875. Also, M(1h) =200e^-0.003875 =200* 0.996132498 = 199.226 grams( on rounding off to the nearest milligram). Thus, the one hour’s change is[(200-199.226)/200]*100 % = 0.387% = 0.39 %( on rounding off to 2 decimal places).

i). Since 1 hour = 1/24 day, so that t would be replaced by h/24 . Hence C = 200e^-0.093*h/24 = 200e^-0.003875, where C is the mass of the remaining Iodine 131 in the sample measured in grams in terms of the number of hours h since the samples mass was originally measured