Question

The half-life of radioactive actinium (^227 Ac) is 22 years. How much will remain after 15 years? Find R first.

Answer 1

Use the formula below.

Let A_0 = 1 because you just want to know the percentage left. Usually that's the initial amount.

t = 15, the number of years past. n = 22, the half-life.

$A=A_0(0.5)^{\frac{t}{n}} \\ \\ A=0.5^{\frac{15}{22}} \\ \\ A=0.62 = 62\%$

Answer 2

62% of the radioactive actinium would be present after 15 years.

The half life of an element is the time taken for the element to decay to half of its initial value. It is given by:

$N(t)=N_o(\frac{1}{2} )^\frac{t}{t_\frac{1}{2} }$

N(t) is the quantity of substance after t days, N₀ is the initial value and t₁₋₂ = half life

Given N₀ = initial value, t = 15 years, t₁₋₂ = 22 years, hence:

$N(15)=N_o*(\frac{1}{2} )^\frac{15}{22 }=0.62N_o$

Therefore 62% of the radioactive actinium would be present after 15 years.

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