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Americium-241 is a radioactive substance used in smoke detectors. The half life of americium is 432 years. If a smoke detector initially contains 1 gram of Americium 241, how much will remain in 432 years?


Question 4 options:


0.5 grams



1.0 grams



1.5 grams



2.0 grams

Verified Answer :

Answer : The correct option is, (2) 0.5 grams

Solution : Given,

As we know that the radioactive decays follow first order kinetics.

First we have to calculate the half life of Americium-241.

Formula used : \(t_{1/2}=\frac{0.693}{k}\)

Putting value of half-life in this formula, we get the rate constant.

\(432years=\frac{0.693}{k}\)

\(k=1.6\times 10^{-3}year^{-1}\)

The expression for rate law for first order kinetics is given by :

\(k=\frac{2.303}{t}\log\frac{a}{a-x}\)

where,

k = rate constant  = \(1.6\times 10^{-3}year^{-1}\)

t = time taken for decay process  = 432 years

a = initial amount of the Americium-241 = 1 g

a - x = amount left after decay process  = ?

Putting values in above equation, we get

\(1.6\times 10^{-3}=\frac{2.303}{432}\log\frac{1}{a-x}\)

\(a-x=0.502g=0.5g\)

Therefore, the amount remain in 432 years will be, 0.5 grams

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