Question

Determine the mass of $N{a^{22}}$ which has an activity of $5mCi$ . Half life of $N{a^{22}}$is $2.6$ years. Avogadro number as $6.023 \times {10^{23}}$ atoms.

Answer 1

Hint: The decay rate of a sample of radioactive material is expressed in terms of curie which is denoted by the symbol $\left( {Ci} \right)$ and a curie is equal to the quantity of radioactive material in which thirty seven billion number of atoms decaying per second i.e. $3.7 \times {10^{10}}{\text{dps}}$, ${\text{dps}}$ stands for disintegrations per second.

Complete step-by-step answer:

Formula used: $A = \dfrac{{0.693}}{{{T_{1/2}}}} \times N$
where $A$ is activity and ${T_{_{1/2}}}$ is half life.
$\dfrac{m}{M} = \dfrac{N}{{{N_A}}}$

where $m$ is the mass of atom, $M$ is the atomic weight of atom, $N$ is the number of atom and ${N_A}$ is the Avogadro's number i.e. $6.023 \times {10^{23}}$atoms.

Given that,
Activity of $N{a^{22}}$ i.e. $A = 5mCi$
as we know that $1Ci{\text{ = }}3.7 \times {10^{10}}{\text{dps}}$
$\therefore 5mCi = 5 \times {10^{ - 3}} \times 3.7 \times {10^{10}} = 1.85 \times {10^8}{\text{dps}}$

And Half life $\left( {{T_{1/2}}} \right)$ of $N{a^{22}}$ $ = 2.6$ years
$\therefore {T_{_{1/2}}} = 2.6 \times 365 \times 86400 = 8.2 \times {10^7}$ seconds

From radioactive decay equation, activity$\left( A \right)$ of radioactive substance is given by the formula:

$A = \dfrac{{0.693}}{{{T_{1/2}}}} \times N$
as $A = 1.85 \times {10^8}{\text{dps}}$, ${T_{_{1/2}}} = 8.2 \times {10^7}$seconds, putting the values we get:
$
  1.85 \times {10^8} = \dfrac{{0.693}}{{8.2 \times {{10}^7}}} \times N \\
   \Rightarrow N = 2.2 \times {10^{16}} \\
$
Now consider the mass of $N{a^{22}}$ $ = m$
and we know that $\dfrac{m}{M} = \dfrac{N}{{{N_A}}}$
where $m$ is the mass of $N{a^{22}}$atom which is to be calculated, $M$ is the atomic weight of $N{a^{22}}$atom i.e. $23$, $N$ is the number of atoms in $N{a^{22}}$ i.e. $2.2 \times {10^{16}}$and ${N_A}$ is Avogadro's number i.e. $6.023 \times {10^{23}}$ atoms.
$\therefore m = \dfrac{{M \times N}}{{{N_A}}}$

Putting the values we get:
$
  m = \dfrac{{23 \times 2.2 \times {{10}^{16}}}}{{6.023 \times {{10}^{23}}}} \\
  m = 8.4 \times {10^{ - 6}} \\
$

Hence the mass of $N{a^{22}}$ which has an activity of $5mCi$ and half life as $2.6$ years is found to be $8.4 \times {10^{ - 6}}$ grams or $8.4$ micrograms.

Note: In this question first we converted the given activity into disintegrations per second and given half life into seconds after that using the radioactive decay equation we calculated the number of $N{a^{22}}$ atoms as $2.2 \times {10^{16}}$. Then we put the mathematical values in the formula $\dfrac{m}{M} = \dfrac{N}{{{N_A}}}$ and after simplifying it we calculated the mass of $N{a^{22}}$ which has an activity of $5mCi$ and half life as $2.6$ years as $8.4$ micrograms.