Calculus Question

Question

The half-life of cesium-137 is 30 years. Suppose we have a 150-mg sample.

(a) Find the mass that remains after t years.

y(t) =  ______ mg

(b) How much of the sample remains after 120 years? (Round your answer to two decimal places.)

_______mg

(c) After how long will only 1 mg remain? (Round your answer to one decimal place.)

t =  _______yr

Answer 1

1/2 = e^30r where r = annual rate of decay

take natural logs of both sides

ln(1/2) =30r

r= ln(1/2)/30 = --.023105

y(t) = 150e^-.023105t where t is measured in years

in 120 years with half life of 30 years that's 120/30 = 4 half lives

or 1/2 of 1/2 of 1/2 of 1/2 = (1/2)^4 = 1/16 x 150= 9.375 mg left

or y =150e(-0123105x120) = 150e^-2.772 = 9.38 mg left after120 years

in one half life 75 is left

2 half lives 37.5

3 half lives 18.75

4 half lives 9.375

5 half lives 4.6875

6 half lives 2.34375

7 half lives 1.671875

8 half lives is less than 1 mg left

7 half lives = 7x30 = 210 years

8 half lives = 8x30 = 240

to get 1 mg left is somewhere in between 210 and 240, maybe the midpoint = 225 years

1 = 150e^-0.0231t, solve for t, take natural logs of both sides

1/150 = e^-0.0231t

ln(1/150) = -.0231t

t = ln(1/150)/-0.0231 = about 216.91 years to reach 1 mg

Answer 2