Chapter 20, Problem 20.98QP

(a)

Interpretation Introduction

Interpretation: The total energy released by the quantity of ${}_{\text{84}}^{\text{210}}\text{Po}$ over the course of 138 days should be calculated& the equation should be drawn for the both cases (a)& (b).

Concept Introduction

• Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

           $\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

• On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.

To calculate the nuclear binding energy.:

• Firstly determine the difference between the mass of the nucleus and the mass of all the protons and neutrons, which yields the mass defect.
• Next, we must apply Einstein mass – energy relationship.
• The nuclear binding energy per nucleon is given by the following equation:
• Nuclear binding energy per nucleus= $\frac{\text{nuclear binding energy}}{\text{number of nucleons}}$
• Einstein mass energy relationship [ $\text{ΔE}=({\text{ΔM)c}}^{\text{2}}$].

Answer to Problem 20.98QP

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

Explanation of Solution

Explanation

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses. Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. So for any nuclear reaction, short hand notation will be in this form that is,

So the equation can be written as, for (a) case is

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

(b)

Interpretation Introduction

Interpretation: The total energy released by the quantity of ${}_{\text{84}}^{\text{210}}\text{Po}$ over the course of 138 days should be calculated& the equation should be drawn for the both cases (a)& (b).

Concept Introduction

• Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

           $\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

• On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.

To calculate the nuclear binding energy.:

• Firstly determine the difference between the mass of the nucleus and the mass of all the protons and neutrons, which yields the mass defect.
• Next, we must apply Einstein mass – energy relationship.
• The nuclear binding energy per nucleon is given by the following equation:
• Nuclear binding energy per nucleus= $\frac{\text{nuclear binding energy}}{\text{number of nucleons}}$
• Einstein mass energy relationship [ $\text{ΔE}=({\text{ΔM)c}}^{\text{2}}$].

Answer to Problem 20.98QP

${}_{\text{84}}^{\text{210}}\text{Po}\stackrel{}{\to }{\text{Pb}}_{\text{82}}^{\text{206}}{\text{+He}}_{\text{2}}^{\text{4}}$

Explanation of Solution

Explanation

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses. Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. So for any nuclear reaction, short hand notation will be in this form that is,

So the equation can be written as, for (a) case is

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

So the equation can be written as, for (b) case is

${}_{\text{84}}^{\text{210}}\text{Po}\stackrel{}{\to }{\text{Pb}}_{\text{82}}^{\text{206}}{\text{+He}}_{\text{2}}^{\text{4}}$

(c)

Interpretation Introduction

Interpretation: The total energy released by the quantity of ${}_{\text{84}}^{\text{210}}\text{Po}$ over the course of 138 days should be calculated& the equation should be drawn for the both cases (a)& (b).

Concept Introduction

• Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

           $\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

• On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.

To calculate the nuclear binding energy.:

• Firstly determine the difference between the mass of the nucleus and the mass of all the protons and neutrons, which yields the mass defect.
• Next, we must apply Einstein mass – energy relationship.
• The nuclear binding energy per nucleon is given by the following equation:
• Nuclear binding energy per nucleus= $\frac{\text{nuclear binding energy}}{\text{number of nucleons}}$
• Einstein mass energy relationship [ $\text{ΔE}=({\text{ΔM)c}}^{\text{2}}$].

To identify:  The energy of an emitting alpha particle

Answer to Problem 20.98QP

The energy of an emitting alpha particle is found to be $\text{E=1}{\text{.03×10}}^{\text{-12}}$ J.

Explanation of Solution

The mass defect ends up as the kinetic energy of the alpha particle

$\text{Δm=}\left(\text{209}\text{.98286amu-205}\text{.97444amu-4}\text{.00150amu}\right)\left(\frac{\text{1kg}}{\text{6}{\text{.022×10}}^{\text{26}}\text{amu}}\right)$

$\text{Δm=1}{\text{.149×10}}^{\text{-29}}\text{kg}$

Einstein mass energy relationship:

${\text{E=Δmc}}^{\text{2}}$

$\text{E=}\left(\text{1}{\text{.149×10}}^{\text{-29}}\text{kg}\right){\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}$

$\text{E =1}{\text{.03×10}}^{\text{-12}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}$

The energy of an emitting alpha particle is found to be

$\text{E=1}{\text{.03×10}}^{\text{-12}}$ J

(d)

Interpretation Introduction

Interpretation: The total energy released by the quantity of ${}_{\text{84}}^{\text{210}}\text{Po}$ over the course of 138 days should be calculated& the equation should be drawn for the both cases (a)& (b).

Concept Introduction

• Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

           $\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

• On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.

To calculate the nuclear binding energy.:

• Firstly determine the difference between the mass of the nucleus and the mass of all the protons and neutrons, which yields the mass defect.
• Next, we must apply Einstein mass – energy relationship.
• The nuclear binding energy per nucleon is given by the following equation:
• Nuclear binding energy per nucleus= $\frac{\text{nuclear binding energy}}{\text{number of nucleons}}$
• Einstein mass energy relationship [ $\text{ΔE}=({\text{ΔM)c}}^{\text{2}}$].

To identify:  The number of polonium atoms in 1 $\text{μ}$ g:

Answer to Problem 20.98QP

The number of polonium atoms in 1 $\text{μ}$ g is found to be $\text{1}{\text{.5×10}}^{\text{3}}\text{=1}\text{.5kJ}$

Explanation of Solution

The number of polonium atoms in 1 $\text{μ}$ g:

$\left({\text{1×10}}^{\text{-6}}\text{Po}\right)\left(\frac{\text{1mol}}{{\text{210g}}^{\text{210}}\text{Po}}\right)\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{atoms}}{\text{1mol}}\right)\text{=2}{\text{.9×10}}^{\text{15}}{}^{\text{210}}\text{Poatoms}$

The half-life of polonium -210 is 138 days.

Half of the original atoms will decay in 138 days. (One half-life period)

The number of alpha particle produce during the 138- day’s period is equal to the number of ${}^{\text{210}}\text{Po}$ atoms that disintegrate.

$\text{(2}{\text{.9×10}}^{\text{15}}{}^{\text{210}}\text{Po atoms) }\left(\frac{\text{1}}{\text{2}}\right)\text{= 1}{\text{.5× 10}}^{\text{15}}{\text{ disintegrated }}^{\text{210}}\text{Po atoms = 1}{\text{.5 ×10}}^{\text{15 }}\text{α particles}$

Using the result from part (c):

So the number of polonium atoms in 1 $\text{μ}$ g is found to be $\text{1}{\text{.5×10}}^{\text{3}}\text{=1}\text{.5kJ}$

$\text{(1}{\text{.5×10}}^{\text{15}}\text{ α particles}\left(\text{1}{\text{.03×10}}^{\text{-12}}\text{J/α particles}\right)\text{=1}{\text{.5×10}}^{\text{3}}\text{=1}\text{.5kJ}$

Conclusion

The total energy released by the quantity of ${}_{\text{84}}^{\text{210}}\text{Po}$ over the course of 138 days has been calculated.