Question

A mixture of $_{{\text{239}}}{\text{Pu}}$and $_{{\text{240}}}{\text{Pu}}$ has a specific activity of ${\text{6}} \times {\text{1}}{{\text{0}}^{\text{9}}}$ dis/s/g. The half-lives of the isotopes are ${\text{2}}{\text{.44}} \times {\text{1}}{{\text{0}}^4}$year and ${\text{6}}{\text{.58}} \times {\text{1}}{{\text{0}}^3}$year respectively. Calculate the isotopic composition of this sample.
A. $_{{\text{239}}}{\text{Pu}}\,{\text{ = }}\,{\text{42}}$%, $_{{\text{240}}}{\text{Pu}}\,{\text{ = }}\,{\text{58}}$%
B. $_{{\text{239}}}{\text{Pu}}\,{\text{ = }}\,{\text{41}}$%, $_{{\text{240}}}{\text{Pu}}\,{\text{ = }}\,{\text{59}}$%
C. $_{{\text{239}}}{\text{Pu}}\,{\text{ = }}\,{\text{40}}$%, $_{{\text{240}}}{\text{Pu}}\,{\text{ = }}\,60$%
D. $_{{\text{239}}}{\text{Pu}}\,{\text{ = }}\,39$%, $_{{\text{240}}}{\text{Pu}}\,{\text{ = }}\,61$%

Answer 1

To determine the answer we should know the formula of specific activity. First we will calculate the specific activity of each isotope. Then by assuming the fraction of one isotope as x and second is as $1 - {\text{x}}$, we will determine the value of x that will give the composition of one isotope of the sample.

Complete step-by-step solution:
First we will convert the half-lives from year to second as follows:
For $_{{\text{239}}}{\text{Pu}}$,
${\text{1year}}\,{\text{ = }}\,{\text{365days}}\,{\ \times }\,\,{24hour \times }\,\,{\text{60}}\,{minute \times }\,\,{\text{60seccond}}$
${\text{2}}{\text{.44}} \times {\text{1}}{{\text{0}}^4}{\text{year}}\, = \,{\text{7}}{\text{.69}} \times {\text{1}}{{\text{0}}^{11}}{\text{second}}$
For $_{{\text{240}}}{\text{Pu}}$,
${\text{6}}{\text{.58}} \times {\text{1}}{{\text{0}}^3}{\text{year}}\, = \,{\text{2}}{\text{.07}} \times {\text{1}}{{\text{0}}^{11}}{\text{second}}$year
The formula which relates the half-life and specific activity is as follows:
${\text{a}}\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}\,{{\text{N}}_{\text{a}}}}}{{{{\text{t}}_{{\text{1/2}}}}{\text{M}}}}$
Where,
${\text{a}}$is the specific activity
${{\text{N}}_{\text{a}}}$is the Avogadro number
${{\text{t}}_{{\text{1/2}}}}$is the half-life
${\text{M}}$is the molar mass
We will calculate the specific activity of each isotope as follows:
On substituting $6.023 \times {10^{23}}$ for Avogadro number, ${\text{7}}{\text{.69}} \times {\text{1}}{{\text{0}}^{11}}$for half-life of $_{{\text{239}}}{\text{Pu}}$, $239$for molar mass of $_{{\text{239}}}{\text{Pu}}$,
${\text{a}}\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}\, \times \,6.023 \times {{10}^{23}}}}{{{\text{7}}{\text{.69}} \times {\text{1}}{{\text{0}}^{11}}\, \times {\text{239}}}}$
${\text{a}}\,{\text{ = }}\,\dfrac{{4.17 \times {{10}^{23}}}}{{{\text{1}}{\text{.83}} \times {\text{1}}{{\text{0}}^4}\,}}$
${\text{a}}\,{\text{ = }}\,2.27 \times {10^9}$
So, the specific activity of $_{{\text{239}}}{\text{Pu}}$is $2.27 \times {10^9}$ /s/g.
On substituting $6.023 \times {10^{23}}$ for Avogadro number, ${\text{2}}{\text{.07}} \times {\text{1}}{{\text{0}}^{11}}$for half-life of $_{{\text{240}}}{\text{Pu}}$, $240$for molar mass of$_{{\text{240}}}{\text{Pu}}$,
$\Rightarrow {\text{a}}\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}\, \times \,6.023 \times {{10}^{23}}}}{{{\text{2}}{\text{.07}} \times {\text{1}}{{\text{0}}^{11}}\, \times {\text{240}}}}$
$\Rightarrow {\text{a}}\,{\text{ = }}\,\dfrac{{4.17 \times {{10}^{23}}}}{{{\text{4}}{\text{.98}} \times {\text{1}}{{\text{0}}^{13}}\,}}$
$\Rightarrow {\text{a}}\,{\text{ = }}\,8.37 \times {10^9}$

So, the specific activity of $_{{\text{240}}}{\text{Pu}}$is $8.37 \times {10^9}$ /s/g.
We assume that amount of $_{{\text{239}}}{\text{Pu}}$is x than the amount of $_{{\text{240}}}{\text{Pu}}$ will be $1 - {\text{x}}$. So, the specific activity of $_{{\text{239}}}{\text{Pu}}$will be $2.27 \times {10^9}$ x and specific activity of $_{{\text{240}}}{\text{Pu}}$ will be $8.37 \times {10^9}$ $1 - {\text{x}}$. It is given that specific activity of the mixture is ${\text{6}} \times {\text{1}}{{\text{0}}^{\text{9}}}$ dis/s/g.

So, the sum of specific activity of $_{{\text{239}}}{\text{Pu}}$and $_{{\text{240}}}{\text{Pu}}$ will be equal to ${\text{6}} \times {\text{1}}{{\text{0}}^{\text{9}}}$ dis/s/g. so,
$\Rightarrow 2.27 \times {10^9}\,{\text{x}}\,{\text{ + }}\,8.37 \times {10^9}\,(1 - {\text{x)}}\,{\text{ = }}\,{\text{6}} \times {\text{1}}{{\text{0}}^{\text{9}}}$
$\Rightarrow 2.27 \times {10^9}\,{\text{x}}\,{\text{ + }}\,8.37 \times {10^9} - 8.37 \times {10^9}{\text{x}}\,{\text{ = }}\,{\text{6}} \times {\text{1}}{{\text{0}}^{\text{9}}}$
$\Rightarrow 2.27 \times {10^9}\,{\text{x}}\, - 8.37 \times {10^9}{\text{x}}\,{\text{ = }}\,{\text{6}} \times {\text{1}}{{\text{0}}^{\text{9}}} - 8.37 \times {10^9}$
$\Rightarrow 2.27 \times {10^9}\,{\text{x}}\, - 8.37 \times {10^9}{\text{x}}\,{\text{ = }}\,{\text{6}} \times {\text{1}}{{\text{0}}^{\text{9}}} - 8.37 \times {10^9}$
$\Rightarrow - 6.1 \times {10^9}{\text{x}}\, = \, - 2.37 \times {10^9}$
$\Rightarrow {\text{x}}\, = \,\dfrac{{2.37 \times {{10}^9}}}{{6.1 \times {{10}^9}}}$
${\text{x}}\, = \,0.39$

So, the fraction of $_{{\text{239}}}{\text{Pu}}$ in the total mixture is $0.39$. We can multiply this fraction with $100$ to convert it into percentage. So,
$\Rightarrow _{{\text{239}}}{\text{Pu}}\,{\text{ = }}\,0.39 \times \,100$
$\Rightarrow _{{\text{239}}}{\text{Pu}}\,{\text{ = }}\,39$%
The fraction of $_{{\text{240}}}{\text{Pu}}$ is $1 - {\text{x}}$ so, on substituting the value of x,
$\Rightarrow _{{\text{240}}}{\text{Pu}}\,{\text{ = }}\,{\text{1}}\, - \,39$
$\Rightarrow _{{\text{240}}}{\text{Pu}}\,{\text{ = }}\,61$%
So, the isotopic composition of the given sample is $_{{\text{239}}}{\text{Pu}}\,{\text{ = }}\,39$%, $_{{\text{240}}}{\text{Pu}}\,{\text{ = }}\,61$%.

Thus, the correct option is (D).

Note: The radioactive disintegration per second per kilogram is known as specific activity. The atoms having the same atomic number but different mass numbers are known as isotopes. The isotopes have the same atomic number hence represent the same element. The total fraction of all isotopes of an element is considered as $100$%.