Question

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

100-mg sample.
(a) Find the mass that remains after years.
(b) How much of the sample remains after 100 years?
(c) After how long will only 1 mg remain?

Answer 1

(a) If
`y(t)` is the mass (in mg) remaining after
`t` years, then
`y(t) = y(0) e^(kt)=100e^(kt).`


`y(30) = 100 e^(30k) = (1)/(2)(100) \implies e^(30k) = (1)/(2) \implies k = -(\ln 2) /30 \implies \\ \\ y(t) = 100e^(-(\ln 2)t/30) = 100 \cdot 2^(-t/30)`

(b)
`y(100) = 100 \cdot 2^(-100/30) \approx \text{9.92 mg}`

(c)

`100 e^(- (\ln 2)t/30) = 1\ \implies\ -(\ln 2) t / 30 = \ln (1)/(100)\ \implies\ \\ \\ t = -30 (\ln 0.01)/(\ln 2) \approx \text{199.3 years}`