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13.9 Obtain the amount of ${}_{27}^{60}{C o}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}{C o}$ is 5.3 years.

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Payal Gupta | Contributor-Level 10

4 months ago

13.9 The strength of the radioactive source is given as:

$\frac{d N}{d t}$ = 8.0 mCi = 8 $\times 10^{- 3}$ $\times 3.7 \times 10^{10}$ decay/s = 296 $\times 10^{6}$ decay/s, where

N = Required number of atoms

Given, half life of ${}_{27}^{60}{C o}$ , $T_{1 / 2}$ = 5.3 years = 5.3 $\times 365 \times 24 \times 60 \times 60$ secs = 167 $\times 10^{6}$ s

For decay constant $λ$ , we have rate of decay as:

$\frac{d N}{d t}$ = $λ N$ or

N = $\frac{1}{λ} \frac{d N}{d t}$ , where $λ$ = $\frac{0.693}{T_{1 / 2}}$ = $\frac{0.693}{167 \times 10^{6}}$ /s = 4.1497 $\times 10^{- 9}$ $s^{- 1}$

N = $\frac{296 \times 10^{6}}{4.1497 \times 10^{- 9}}$ = 7.133 $\times 10^{16}$ atoms

For ${}_{27}^{60}{C o}$ , mass of 6.023 $\times 10^{23}$ atoms = 60 gms

Therefore, the mass of 7.133 $\times 10^{16}$ atoms = $\frac{60}{6.023 \times 10^{23}} \times$ 7.133 $\times 10^{16}$ gms = 7.106 $\times 10^{- 6}$ g

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13.9 Obtain the amount of ${}_{27}^{60}{C o}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}{C o}$ is 5.3 years.

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13.9 Obtain the amount of ${}_{27}^{60}{C o}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}{C o}$ is 5.3 years.