After the sudden release of radioactivity from the Chernobylnuclear reactor acci
ID: 1756853 • Letter: A
Question
After the sudden release of radioactivity from the Chernobylnuclear reactor accident in 1986, the radioactivity of milk inPoland raised to 2,000 Bq/L due to iodine 131, with a half life of8.04 days. Radioactive iodine is particularly hazardous because thethyroid gland concentrates iodine. The Chernobyl accident caused ameasurable increase in thyroid cancers among children in Belarus.a.) For comparison, find the activity of milk due to potassium.Assume 1 liter of milk contains 2.00g of potassium. of which 0.0117% is the isotope 40K, which has a half-life of 1.28 x10^9 years. b.) After what length of time would the activity due toiodine fall below that due to potassium?I don't remember what I got for part A but for part B I got 1.31years and that is showing up as incorrect. Can anyone help me withthis question? Thanks!
Explanation / Answer
Given : a)= ln(2) / half life
= ln(2)/1.28e9 yr
= 5.41521235*10-10 per year
N = 2 g *(6.022e23/39.10 g) *(0.000117) =5.22449604*1018 atoms
Activity = *N
= 5.22449604*1018 atoms*5.41521235*10-10 peryear
= 2.82917555e9 disintegrations per year.
2.82917555*109 disintegration/year *1year/365 days * 1day /24hr * 1hr/60 min* 1 min/60s
= 89.7125682 disintegration/second Note 1disintegration /second = 1 Bq
Activity = Initial Activity*exp(-*t)
_Iodine = ln(2)/8.04 days = 0.0862123359 per day
Activity of Iodine: 2000 Bq *exp(-6.4501196e-5*t)
_Potassium = ln(2)/1.28e9 yr *(1yr/365 day) = 1.48361982e-12per day
Activity of Potassium = 89.7125682*exp(-1.48361982e-12*t)
Activity Iodine = Activity of Potassium
2000 *exp(-6.4501196e-5*t) =89.7125682*exp(-1.48361982e-12*t)
Solve for t
exp(-6.4501196e-5*t + 1.48361982e-12*t) = 89.7125682/2000
exp(-6.45011945e-5*t) = 89.7125682/2000
t = ln(89.7125682/2000) / -6.45011945e-5
t = 48127.6604 days
t = 132 years
I hope it helps you