Question

The half-life of cesium-137 is 30 years. Suppose we have a 18-gram sample.

(a) Find the yearly growth factor a. (Round your answer to five decimal places.)

a =

(b) Find an exponential model m(t) = Cat for the mass remaining after t years.

m(t) =

(c) How much of the sample will remain after 85 years? (Round your answer to two decimal places.)

g

(d) After how long will only 3 g of the sample remain? (Round your answer to one decimal place.)

yr

Answer 1

Sol amount of cesium_134 at a time į het Net) be the By Natural law ON JE OM in grauth contact da = -adt whee à Ida --afde In en) = -at tince) hocalace) =-at . ln (~) = -at Ņe at NECě a I whe – i integration ental-

Emitally, we have 18 game of sample re. N= 18, when t=0 form equitan o 18 = 000 =116=18) IN(E) = 1889). Half life of Cenium - 137 is 30 years =) N=1 when t = 30 9=18 e 300 0209 -302 = ench) a= ln(k)=-(-0-6931471806) P =0.02310490602 Tan 0.02310) LT -02306 How I met = 18e at sige, TNCE ) = m(t) say) © Now to fd mee), bohen t = 85 m(L) = 18 20.02310185) mit ) = 0.140366x 18 imeyty gramce) = 2.30 am Now to find t=? when me) = 3 @ 3 =10.023101 -0-023106 g -0.02310 t = ln (y) tas (ln(65) = 77.5653 TE = 47.67 year 0.02310