Question

Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 40.0 % of an Am-241 sample to decay? Express your answer with the appropriate units. View Available Hint(s) t t t = nothing nothing

Answer 1

Answer:

319 years

Explanation:

For a radioactive decay we have

N = N₀e^-kt , where:

N= Particles  Remainng after a time t

N₀ = Particles  Initially  present

k = ln 2/ t₁/₂ , t₁/₂ is the half life of the radioactive element (Am-241)

We are given that the half-life is 432 yrs, from this inhformation we can calculate the value of  k which then will be used to calculate the time Am-241 will take to decay to 40% once we realize we are given the ratio N/N₀.

k = ln 2/ 432 yr = 0.693/ 432 yr   = 1.6 x 10⁻³ /yr

N/N₀ =  e^-kt  

N/N₀ = 60 (amount remaining after 40 % has decayed)/100

N/N₀ = 0.60

0.60 = e-^-kt

taking natural log to both sides of the equation to get rid of e:

ln (0.60) = -1.6 x 10⁻³ /yr x  t    ∴   t = - ln (0.60) /1.6 x 10⁻³/ yr

t = 0.51 /1.6 x 10⁻³ yr = 319 yrs

To verify our answer realize that what is being asked is how many years it will take to decay 40 %, and we are told the half life , 50 % decay , is 432 years, so for 40 % we will expect it will take less than that which agrees with our resul of 319 years.