Question 1023633
a.1/2=e^-30.2r
-ln 2=ln e^-30.2r
-0.69314718055994530941723212145818=ln e^-30.2r=-30.2r
r=0.02295189339602467911977589806153
The rate of decay constant is 0.022951893. 2011+30.2=2041 as the year in which half of the cesium is gone.

b.The amount of cesium left after t is 1/4. So:
ln 1/4=-0.022951893t
ln 1/4=-ln 4=-1.3862943611198906188344642429164
t=-1.3862943611198906188344642429164/-0.022951893=60.40000 years.
2011+60.4=2071 as the year cesium levels are 1/4 of what they were in 2011.

c.Since the reactor released 4.2 times the amount released in Chernobyl, we have:
ln 1/4.2=-0.022951893t
-ln 4.2=-0.022951893t
t=62.525758781 years after 2011, or 2073.52 when Fukushima levels equal Chernobyl levels.