What is the activity of one gram of ${}_{88}^{226}{Ra},$ whose half-life is 1622 years?

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Solution

The number of atoms in 1g of radium is,

N = \frac{6.023 \times 10^{23}}{226} = 2.666 \times 10^{21} atoms

The decay constant is related to the half-life by,

λ = \frac{0.693}{T_{1 / 2}}

Therefore,

λ = (\frac{0.693}{1622 y}) (\frac{1 y}{365 d}) (\frac{1 d}{8.64 \times 10^{4} s}) = 1.355 \times 10^{11} s^{- 1}

The activity is then found from,

Activity = λN = (1.355 \times 10^{- 11} s^{- 1}) (2.666 \times 10^{21}) = 3.612 \times 10^{10} disintegration / s .

The definition of the curie is 1 Ci = 3.7 × 10¹⁰ disintegrations/s. This is approximately equal to the value found above.

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