what is the half-life of this isotope?

Question

²⁴¹₉₅Am  is used in many home smoke alarms. If 85% of the americium in a smoke detector decays in 1250 years, what is the half-life of this isotope?

The correct answer is 4.57 x 10² yr

Answer 1

Another approach would be the following:

Use the expression: Fraction remaining = 0.5ⁿ where n = number of half lives elapsed

If 85% decays, then fraction remaining = 15% = 0.15 as a fraction

Solve for n (# of half lives elapsed)

0.15 = 0.5n

log 0.15 = n log 0.5

n = 2.737 half lives

So, 2.737 half lives = 1250 yrs. Using a simple proportion, we can solve for the value of 1 half life...

2.737 half lives/1250 yr = 1 half life/x yrs

x = 4.57x10² yrs

Answer 2

The equation for half life can be written as:

a(t) = a₀(1/2)^t/k where a₀ is the initial amount, "k" is the half life in years, and "t" is the number of years that have gone by. For example, if the half life of a material is 10 years and you started with 100 grams then after 20 years you would calculate the amount of material left as:

a(20) = 100(1/2)^20/10 = 100(1/2)² = 100(1/4) = 25 grams

So in this case, we can arbitrarily pick an initial amount of 100 grams and then say the amount left after 1250 years is 15 grams (85% is gone) . Then:

a(t) = a₀(1/2)^t/k

15 = 100(1/2)^1250/k

15/100 = (1/2)^1250/k

0.15 = (1/2)^1250/k

log(0.15) = log(1/2)^1250/k

log(0.15) = (1250/k)log(1/2)

log(0.15)/log(1/2) = 1250/k

2.7369656 = 1250/k

k = 1250/2.7369656 = 456.71 years (which is 4.5671 x 10² years)