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$8.0 mCi$ सक्रियता का रेडियोऐक्टिव स्तोत प्राप्त करने के लिए ${ }_{27}^{60} Co$ की कितनी मात्रा की आवश्यकता होगी? ${ }_{27}^{60} Co$ की अर्धायु $5.3$ वर्ष है।
$7.216 \times 10^{-8}\; g$
$1.624 \times 10^{-7}\; g$
$5.162 \times 10^{-5}\; g$
$7.106 \times 10^{-6}\; g$
Solution
The strength of the radioactive source is given as
$\frac{d N}{d t}=8.0 mCi$
$=8 \times 10^{-3} \times 3.7 \times 10^{10}$
$=29.6 \times 10^{7}$ decay $/ s$
Where, $N =$ Required number of atoms
Half-life of $\frac{60}{27} Co , T_{1 / 2}=5.3$ years
$=5.3 \times 365 \times 24 \times 60 \times 60$
$=1.67 \times 108 s$
For decay constant $\lambda,$ we have the rate of decay as $\frac{d N}{d t}=\lambda N$
Where $\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{1.67 \times 10^{8}} s^{-1}$
$\therefore N=\frac{1}{\lambda} \frac{d N}{d t}$
$=\frac{29.6 \times 10^{7}}{\frac{0.693}{1.67 \times 10^{8}}}=7.133 \times 10^{16}$ atoms
For $_{27} Co ^{60}$
Mass of $6.023 \times 1023$ (Avogadro's number) atoms $=60 g$
Mass of $7.133 \times 10^{16}$ atoms $=\frac{60 \times 7.133 \times 10^{16}}{6.023 \times 10^{33}}=7.106 \times 10^{-6} g$
Hence, the amount of $_{27} Co ^{60}$ necessary for the purpose is $7.106 \times 10^{-6}\; g$
