Obtain the amount of $_{27}^{60} Co$ necessary to provide a radioactive source of $8.0\; mCi$ strength. The half-life of $^{60}_{27} Co$ is $5.3$ years.
Obtain the amount of $_{27}^{60} Co$ necessary to provide a radioactive source of $8.0\; mCi$ strength. The half-life of $^{60}_{27} Co$ is $5.3$ years.
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 $7.106 \times 10^{-6}\; g$
Similar Questions
Number of nuclei of a radioactive substance at time $t = 0$ are $1000$ and $900$ at time $t = 2$ $s$. Then number of nuclei at time $t = 4$ $s$ will be
Number of nuclei of a radioactive substance at time $t = 0$ are $1000$ and $900$ at time $t = 2$ $s$. Then number of nuclei at time $t = 4$ $s$ will be
The relation between $\lambda $ and $({T_{1/2}})$ is (${T_{1/2}}=$ half life, $\lambda=$ decay constant)
The relation between $\lambda $ and $({T_{1/2}})$ is (${T_{1/2}}=$ half life, $\lambda=$ decay constant)
- [AIPMT 2000]
A radio isotope $X$ with a half-life $1.4 \times 10^{9}\; years$ decays of $Y$ which is stable. A sample of the rock from a cave was found to contain $X$ and $Y$ in the ratio $1: 7$. The age of the rock is ........ $ \times 10^9\; years$
A radio isotope $X$ with a half-life $1.4 \times 10^{9}\; years$ decays of $Y$ which is stable. A sample of the rock from a cave was found to contain $X$ and $Y$ in the ratio $1: 7$. The age of the rock is ........ $ \times 10^9\; years$
- [AIPMT 2014]
Why radioactivity is considered a nuclear phenomenon ?
Why radioactivity is considered a nuclear phenomenon ?
The normal activity of living carbon-containing matter is found to be about $15$ decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive $_{6}^{14} C$ present with the stable carbon isotope $_{6}^{12} C$. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ($5730$ years) of $_{6}^{14} C ,$ and the measured activity, the age of the specimen can be approximately estimated. This is the principle of $_{6}^{14} C$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of $9$ decays per minute per gram of carbon. Estimate the approximate age (in $years$) of the Indus-Valley civilisation
The normal activity of living carbon-containing matter is found to be about $15$ decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive $_{6}^{14} C$ present with the stable carbon isotope $_{6}^{12} C$. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ($5730$ years) of $_{6}^{14} C ,$ and the measured activity, the age of the specimen can be approximately estimated. This is the principle of $_{6}^{14} C$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of $9$ decays per minute per gram of carbon. Estimate the approximate age (in $years$) of the Indus-Valley civilisation