Chapter 20, Problem 102AP

Interpretation Introduction

Interpretation:

The equation for decay and formation of ${}^{210}\text{Po}$

are to be written. The energy of alpha particle and the energy released from the given amount of ${}^{210}\text{Po}$

are to be determined.

Concept introduction:

In a balanced radioactive decay,

${\displaystyle \sum \text{Reactant mass number = }{\displaystyle \sum \text{Product mass number}}}$

And,

${\displaystyle \sum \text{Reactant atomic number = }{\displaystyle \sum \text{Product atomic number}}}$

The energy during the decay of a radioactive substance is given as:

$\Delta \text{E = (}\Delta {\text{m)c}}^{\text{2}}$

Here, $\Delta \text{E}$

is the amount of energy released, $\Delta \text{m}$

is the mass defect which occurs during the course of decay and c is the speed of the light.

The mass defect ($\Delta \text{m}$

) is given as the difference of total atomic mass of the product and the reactant in the balanced radioactive reaction.

Answer to Problem 102AP

Solution:

(a) ${{}^{209}}_{83}{\text{Bi + }}^{\text{1}}{}_{\text{0}}\text{n }\to {}^{\text{210}}{}_{\text{84}}{\text{Po + }}^{\text{0}}{}_{\text{-1}}\text{β}$

(b) ${{}^{210}}_{84}\text{Po }\to {}^{206}{}_{\text{82}}{\text{Pb + }}^4{}_2\text{He}$

(c) $\text{1}\text{.03 }\times {\text{ 10}}^{-12}\text{ J}$

(d) $\text{1}\text{.5 kJ}$

Explanation of Solution

a) The equation for the reaction; ${}^{210}\text{Po}$

is prepared by bombarding ${}^{210}\text{Bi}$

with neutrons.

It is known that the formation of ${}^{210}\text{Po}$ can be written as:

${{}^{209}}_{83}{\text{Bi + }}^{\text{1}}{}_{\text{0}}\text{n }\to {}^{\text{210}}{}_{\text{84}}\text{Po + X}$

Now, apply the balancing rules as follows:

${\displaystyle \sum \text{Reactant mass number = }{\displaystyle \sum \text{Product mass number}}}$

And

${\displaystyle \sum \text{Reactant atomic number = }{\displaystyle \sum \text{Product atomic number}}}$

On the left hand side of the reaction, sum of atomic number is calculated as follows:

$83+0=83$

On the right–hand side, the sum of atomic number is $\text{atomic number of X }+84$

Thus, Atomic number of X is calculated as follows:

$83-84=-1$

Similarly,

On the left–hand side of the reaction, sum of mass number is calculated as follows;

$209+1=210$

On the right side of the reaction, sum of mass number is $\text{mass number of X}+\text{ 210}$

Thus, mass number of X is calculated as follows:

$210-210=0$

So, X will be ${{}^0}_{-1}\text{β}$

Hence, the balanced reaction is as follows:

${{}^{209}}_{83}{\text{Bi + }}^{\text{1}}{}_{\text{0}}\text{n }\to {}^{\text{210}}{}_{\text{84}}{\text{Po + }}^{\text{0}}{}_{\text{-1}}\text{β}$

b) The equation of decay process; ${}^{210}\text{Po}$ decays with alpha particle emission.

It is known that the decay of ${}^{210}\text{Po}$ can be written as follows:

${{}^{210}}_{84}\text{Po }\to {\text{ X + }}^4{}_2\text{He}$

Now, apply the balancing rules as follows:

${\displaystyle \sum \text{Reactant mass number = }{\displaystyle \sum \text{Product mass number}}}$

And

${\displaystyle \sum \text{Reactant atomic number = }{\displaystyle \sum \text{Product atomic number}}}$

On the left-hand side of the reaction, sum of atomic number is $84$

On the right-hand side, the sum of atomic number is $\text{atomic number of X }+\text{2}$

Thus, atomic number of X is calculated as follows:

$84-2=82$

Similarly,

On the left hand side of the reaction, sum of mass number is $210$

On the right side of the reaction, sum of mass number is $\text{mass number of X}+\text{ 4}$

Thus, mass number of X is calculated as follows:

$210-4=206$

So, X will be ${{}^{206}}_{82}\text{Po}$

Hence, the balanced reaction is as follows:

${{}^{210}}_{84}\text{Po }\to {}^{206}{}_{\text{82}}{\text{Pb + }}^4{}_2\text{He}$

c) The energy of an emitted $\alpha$

particle in given reaction:

${{}^{210}}_{84}\text{Po }\to {}^{206}{}_{\text{82}}{\text{Pb + }}^4{}_2\alpha$

Mass of ${}^{210}\text{Po}$

is $\text{209}\text{.98286 amu}$

Mass of ${}^{206}\text{Pb}$

is $\text{205}\text{.97444 amu}$

Mass of ${{}^4}_2\alpha$

is $\text{4}\text{.00150 amu}$

The balanced reaction is given as follows:

${{}^{210}}_{84}\text{Po }\to {}^{206}{}_{\text{82}}{\text{Pb + }}^4{}_2\alpha$

Now, the mass defect $\Delta \text{m}$

is calculated as follows:

$\Delta m=\text{Mass of products}-\text{Mass of reactants}$

Substitute the values of masses in the above expression,

$\begin{array}{c}\Delta m={\text{[ mass of }}^{\text{206}}{\text{Pb + mass of }}^{\text{4}}\alpha \text{]}-[{\text{mass of }}^{210}\text{Pb]}\\ \text{ = [ 205}\text{.97444 amu + }4.00150\text{ amu ]}-[\text{209}\text{.98286 amu ]}\\ \text{ = }-\text{0}\text{.00692 amu}\end{array}$

Now, the magnitude of mass defect is taken for further calculation as mass is converted to energy.

Now, it is known that $1\text{ kg}$ is equivalent to $6.022\times {\text{ 10}}^{\text{26}}\text{amu}$

Thus, the conversion factor is $\frac{1\text{Kg}}{6.022\times {\text{ 10}}^{\text{26}}\text{ amu}}$

Hence, the mass defect can be converted to kilogram unit by using the above conversion factor as follows:

$\begin{array}{l}\Delta m=\left(0.00692\cancel{\text{amu}}\times \frac{1\text{kg}}{6.022\times {\text{ 10}}^{\text{26}}\cancel{\text{amu}}}\right)\\ \text{ = 1}\text{.149 }\times {\text{10}}^{-\text{29}}\text{kg}\end{array}$

The energy during the decay of a radioactive substance is given as follows:

$\Delta E=(\Delta m){\text{c}}^{\text{2}}$

Substitute the values of energy and mass defect in the above expression,

$\begin{array}{c}\Delta \text{E = (1}\text{.149}\times {\text{10}}^{-\text{29}}\text{kg)(3}\times {\text{10}}^{\text{8}}{\text{m/sec)}}^{\text{2}}\\ \text{ = 1}\text{.03}\times {\text{10}}^{-\text{12}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ =\text{1}\text{.03}\times {\text{10}}^{-\text{12}}\text{ J}\end{array}$

Here, it is assumed that all the energy generated by the mass defect is contributed to the kinetic energy of the alpha particle.

Thus, the kinetic energy of alpha particle is $\text{1}\text{.03 }\times {\text{ 10}}^{\text{-12}}\text{J}$

d) The total energy released by $1\mu \text{g}$

of ${}^{210}\text{Po}$ over the course of 138 days.

At first, calculate the number of ${}^{210}\text{Po}$

atoms in $1\mu \text{g}$.

It is known that one micro gram of any substance is equivalent to ${10}^{-6}\text{g}$

Thus, the conversion factor is $\left(\frac{{10}^{-6}{\text{ g }}^{\text{210}}\text{Po}}{1\mu \text{g}}\right)$.

It is known that one mol of any substance is equivalent to its molar mass.

Thus, the conversion factor for ${}^{210}\text{Po}$

is $\left(\frac{1\text{ mol}}{210{\text{ g }}^{\text{210}}\text{Po}}\right)$.

Similarly, it is known that one mol of any substance is equivalent to Avogadro’s number.

Thus, the conversion factor is $\left(\frac{6.022\times {\text{ 10}}^{\text{23}}\text{ atoms}}{1\text{ mol}}\right)$

Hence, using the appropriate conversion factor, the number of ${}^{210}\text{Po}$

atoms in $1\mu \text{g}$ is calculated as follows:

$\left(\frac{{10}^{-6}\cancel{{\text{g }}^{\text{210}}\text{Po}}}{1\mu \text{g}}\right)\left(\frac{\cancel{1\text{ mol}}}{210\cancel{{\text{g }}^{\text{210}}\text{Po}}}\right)\left(\frac{6.022\times {\text{ 10}}^{\text{23}}\text{ atoms}}{\cancel{1\text{ mol}}}\right)=\text{ 2}\text{.9 }\times {\text{ 10}}^{\text{15}}{}^{\text{210}}\text{Po atoms}$

Now, as per the given information, it is known that half –life of polonium - $210$

is $138\text{ days}$, and therefore it can be said that in $138\text{ days}$, half of polonium-$210$

will decay to form alpha particle. Thus, the number of polonium atoms decayed will be equivalent to the number of alpha particles formed.

Hence, the number of polonium atoms decayed in $138\text{ days}$ is calculated as follows:

$\begin{array}{c}(2.9\times {\text{ 10}}^{\text{15}}{}^{\text{210}}\text{Po atoms)(}\frac{1}{2})=1.5\times {10}^{15}{\text{ disintegrated }}^{\text{210}}\text{Po atoms}\\ \text{ = }1.5\times {10}^{15}\text{ alpha particles}\\ \end{array}$

Now, use the energy result obtained in part (c) to get the energy of the alpha particle as follows:

$\begin{array}{c}\text{Energy}=\left(1.5\times {10}^{15}\cancel{\text{alpha particles}}\right)\left(\text{1}\text{.03 }\times {\text{10}}^{-\text{12}}\text{J/}\cancel{\text{alpha particle}}\right)\\ =\text{1}\text{.5 }\times {10}^3\text{ J }\\ =\text{1}\text{.5 kJ (as, 1 kJ = 1000 J)}\end{array}$

Thus, the total energy released from $1\mu \text{g}$

of polonium is $1.5\text{ kJ}$