Question

1. The Fukushima nuclear reactor accident caused the environmental release of significant amounts of radioactive caesium (Cs) isotopes. Specifically, caesium-134 and caesium- 137, which have half-lives of, respectively, two years and 30 years. The survey of soil radioactivity conducted in June 2011 revealed soil surface Cs-134 activity of about 500,000 Bq/m2 in the town of Okuma (roughly 10 km from the stricken power plant) and Cs-137 activity of roughly the same level. Disregarding any other changes or cleanup efforts since then,

2. a) In what year can the level of soil radioactivity due to these two radioisotopes

   (combined) be expected to have dropped to 50% of its original value?

3. b) How much time (roughly) has to pass before the soil radioactivity due to both

   isotopes (combined) drops to 1% of its original value?

4. c) What would you expect the radioactivity due to Cs-134 to be in Okuma today compared to its original level in June 2011? (Semi-quantitative statements are sufficient.) Explain.

Answer 1

Given ,

Cs-134 half life=2 year

Cs-137 half life=30years

Measured value of radioactivity in June 2011=500000Bq/m^2

1)so for values of radioactive isotope to come to 50% time needed is = half-life period

So the Cs 134 will be at 50% in year 2013

Cs 137 will be 50% in year 2041

2)for radioactive amount to reduce to 1% .

If we have initial value N0, recent value should be N= 0.01 N0

So age = (1/decay const)×ln(NO/N)

For Cs-134 , t=(1/(0.693/2))×ln(N0/N)=2.886×ln(100)=23.29 years

For Ca 237, t=(1/0.693/30)ln(100)=199.358 years

3)if we take today as 2020 June , then time passed t=9 years

So for Cs134 , NO/N = e^(decay constant×t)=e^((0.693/2)×9)=22.61

So N=NO/22.61=0.0442 NO

For Cs137, NO/N= e^((0.693/30)9)=1.231

N=0.8122 N0