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

7 months ago

13.9 The strength of the radioactive source is given as:

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

N = Required number of atoms

Given, half life of ${}_{27}^{60}{C o}$ , $T_{1 / 2}$ = 5.3 years = 5.3 $* 365 * 24 * 60 * 60$ secs = 167 $* 10^{6}$ 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 $λ$ = $\frac{0.693}{T_{1 / 2}}$ = $\frac{0.693}{167 * 10^{6}}$ /s = 4.1497 $* 10^{- 9}$ $s^{- 1}$

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

For ${}_{27}^{60}{C o}$ , ma

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13.9 Obtain the amount of Co2760 necessary to provide a radioactive source of 8.0 mCi strength. The half-life of Co2760 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.