The specific activity of a preparation consisting of radioactive $C o^{58}$ and nonradioactive $C o^{59}$ is equal to $2.2 \cdot 1 0^{12}$ dis/(s $\cdot$ g). The half life of $C o^{58}$ is $71 / 3$ days. Find the ratio of the mass of radioactive cobalt in that preparation to the total mass of the preparation (in per cent).

Question

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Answer 1

We see that
Specific activity of the sample
$= \frac{1}{M + M ^{'}}$ [ Activity of $M g m$ of $C o^{58}$ in the sample]

Here $M$ and $M^{'}$ are the masses of $C o^{58}$ and $C o^{59}$ in the sample. Now activity of $M g m$ of $C o^{58}$

$= \frac{M}{5 8} \times 6.023 \times 1 0^{23} \times \frac{l n 2}{7 1 . 3 \times 8 6 4 0 0}$ dis/sec

$= 1.168 \times 1 0^{15} M$

Thus from the problem
$1.168 \times 1 0^{15} \frac{M}{M + M ^{'}} = 2.2 \times 1 0^{12}$

or $\frac{M}{M + M ^{'}} = 1.88 \times 1 0^{- 3}$ i.e., $0.188 %$

Answer 2

We see that
Specific activity of the sample

= \frac{1}{M + M ^{'}}

[ Activity of

M g m

of

C o^{58}

in the sample]

Here

M

and

M^{'}

are the masses of

C o^{58}

and

C o^{59}

in the sample. Now activity of

M g m

of

C o^{58}

= \frac{M}{5 8} \times 6.023 \times 1 0^{23} \times \frac{l n 2}{7 1 . 3 \times 8 6 4 0 0}

dis/sec

= 1.168 \times 1 0^{15} M

Thus from the problem

1.168 \times 1 0^{15} \frac{M}{M + M ^{'}} = 2.2 \times 1 0^{12}

or

\frac{M}{M + M ^{'}} = 1.88 \times 1 0^{- 3}

i.e.,

0.188 %