Chapter 21, Problem 21.20QP

Interpretation Introduction

Interpretation:

The nuclear binding energy and the binding energy per nucleon for the following isotope s should be calculated

Concept introduction:

• Nuclear binding energy: It is the minimum amount of energy required to disassemble the nucleus of an atom into its component parts.

  The component parts are neutrons and protons, which are collectively called as nucleons.

• Binding energy per nucleon:
• The maximum binding energy per nucleon occurs at around mass number A=50.

  Example –Iron nucleolus (${\text{Fe}}^{56}$) is located closed to the peak with a binding energy per nucleon value of approximately 8.8 MeV.

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,$\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

Formula:

Einstein mass energy relationship [$\text{ΔE}=({\text{ΔM)c}}^{\text{2}}$]

$\text{nuclear binding energy per nucleon =}\frac{\text{nuclear binding energy}}{\text{number of nucleons}}$

Answer to Problem 21.20QP

Thus the nuclear binding energy is $\text{2}{\text{.3603×10}}^{\text{-10}}\text{J}$.

The nuclear binding energy per nucleon is $\text{1}{\text{.2828×10}}^{\text{-12}}\text{J/nucleon}$

Explanation of Solution

The binding energy is the energy required for the process

${\text{He}}_{\text{2}}^{\text{4}}\stackrel{}{\to }{\text{2p}}_{\text{1}}^{\text{1}}{\text{+2n}}_{\text{0}}^{\text{1}}$

There are two proton and 2 neutron in the helium nucleus.

The mass of 2 proton is

$\text{2×1}\text{.00728amu=2}\text{.01456amu}$

The mass of 2 electron is

$\text{2×5}{\text{.4858×10}}^{\text{-4}}\text{=0}\text{.0010972amu}$

The mass of 2 neutron is

$\text{2×1}\text{.008665amu=2}\text{.017330amu}$

Therefore, the predicted mass of helium molecule is

$\text{2}\text{.01456amu+0}\text{.0010972amu+2}\text{.017330amu=4}\text{.03299amu}$

So the mass defect is found to be

$\text{ΔM=4}\text{.002603amu-4}\text{.03299amu=-0}\text{.03039amu}$

$\text{-0}\text{.03039amu×}\frac{\text{1kg}}{\text{6}{\text{.0221418×10}}^{\text{26}}\text{amu}}\text{=-5}{\text{.046×10}}^{\text{-29}}$

The energy change ($\text{ΔE}$) for the process is

$\text{ΔE=(-5}{\text{.046×10}}^{\text{-29}}\text{kg)(2}{\text{.99792458×10}}^{\text{8}}{\text{m/s)}}^{\text{2}}$

$\text{ΔE=-4}{\text{.535×10}}^{\text{-12}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{=-4}{\text{.535×10}}^{\text{-12}}\text{J}$

The nuclear binding energy is $\text{4}\text{.535}\times {\text{10 }}^{-\text{12}}\text{ J}$.

So the energy required to break up one helium nucleus into proton and two neutrons.

For helium nucleus

The nuclear binding energy per nucleon=$\frac{\text{4}\text{.535}\times {\text{10 }}^{-\text{12}}\text{ J}}{\text{4 nucleons}}\text{=1}{\text{.134×10}}^{\text{-12}}\text{J}$

The binding energy is the energy required for the process

${\text{W}}_{\text{74}}^{\text{184}}\to {\text{74}}_{\text{1}}^{\text{1}}{\text{p +110}}_{\text{0}}^{\text{1}}\text{n}$

There are 74 protons and 110 neutrons in the tungsten nucleus.

The mass of 74 proton is

$\text{74 }\times 1.00728amu=74.5387amu$

The mass of 110 neutron is

$\text{74×5}{\text{.4858×10}}^{\text{-4}}\text{amu=0}\text{.04059amu}$

$\text{110×1}\text{.008665amu=110}\text{.9532amu}$

So the predicted mass of ${\text{W}}_{\text{74}}^{\text{184}}$ is

$\text{74}\text{.5387amu+0}\text{.04059amu+110}\text{.9532amu=185}\text{.5325 amu}$

The mass defect is found

$\text{ΔE=183}\text{.950928amu-185}\text{.5325amu=-1}\text{.5816amu}$

$\text{-1}\text{.5816amu×}\frac{\text{1kg}}{\text{6}{\text{.0221418×10}}^{\text{26}}\text{amu}}{\text{=-26262×10}}^{\text{-27}}\text{kg}$

The energy change for the process is 

$\text{ΔE=(-2}{\text{.6262×10}}^{\text{-27}}\text{kg)(2}{\text{.99792458×10}}^{\text{8}}{\text{m/s)}}^{\text{2}}$

$\text{ΔE=-2}{\text{.3603×10}}^{\text{-10}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{=-2}{\text{.3603×10}}^{\text{-10}}\text{J}$

Thus the nuclear binding energy is $\text{2}{\text{.3603×10}}^{\text{-10}}\text{J}$.

The nuclear binding energy per nucleon is obtained as follow:

$\frac{\text{2}{\text{.3603×10}}^{\text{-10}}\text{J}}{\text{184nucleons}}\text{=1}{\text{.2828×10}}^{\text{-12}}\text{J/nucleon}$

To calculate the binding energy and nuclear binding energy per nucleon by considering the given value

The binding energy is the energy required for the process

${\text{He}}_{\text{2}}^{\text{4}}\stackrel{}{\to }{\text{2p}}_{\text{1}}^{\text{1}}{\text{+2n}}_{\text{0}}^{\text{1}}$

There are two protons and 2 neutrons in the helium nucleus.

The mass of 2 protons is

$\text{2×1}\text{.00728amu=2}\text{.01456amu}$

The mass of 2 electrons is

$\text{2×5}{\text{.4858×10}}^{\text{-4}}\text{=0}\text{.0010972amu}$

The mass of 2 neutrons is

$\text{2×1}\text{.008665amu=2}\text{.017330amu}$

Therefore, the predicted mass of helium molecule is

$\text{2}\text{.01456amu+0}\text{.0010972amu+2}\text{.017330amu=4}\text{.03299amu}$

So the mass defect is found to be

$\text{ΔM=4}\text{.002603amu-4}\text{.03299amu=-0}\text{.03039amu}$

$\text{-0}\text{.03039amu×}\frac{\text{1kg}}{\text{6}{\text{.0221418×10}}^{\text{26}}\text{amu}}\text{=-5}{\text{.046×10}}^{\text{-29}}$

The energy change ($\text{ΔE}$) for the process is 

$\text{ΔE=(-5}{\text{.046×10}}^{\text{-29}}\text{kg)(2}{\text{.99792458×10}}^{\text{8}}{\text{m/s)}}^{\text{2}}$

$\text{ΔE=-4}{\text{.535×10}}^{\text{-12}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{=-4}{\text{.535×10}}^{\text{-12}}\text{J}$

The nuclear binding energy is $\text{4}\text{.535}\times {\text{10 }}^{-\text{12}}\text{ J}$.

So the energy required to break up one helium nucleus into proton and two neutrons.

For helium nucleus

The nuclear binding energy per nucleon=$\frac{\text{4}\text{.535}\times {\text{10 }}^{-\text{12}}\text{ J}}{\text{4 nucleons}}\text{=1}{\text{.134×10}}^{\text{-12}}\text{J}$

Formula:

Einstein mass energy relationship [$\text{ΔE}=({\text{ΔM)c}}^{\text{2}}$]

$\text{nuclear binding energy per nucleon =}\frac{\text{nuclear binding energy}}{\text{number of nucleons}}$

Given:

${\text{He}}_{\text{2}}^{\text{4}}$=4.002603amu

${\text{W}}_{\text{74}}^{\text{184}}$=183.950928amu

To calculate the binding energy, and the binding energy per nucleons

The binding energy is the energy required for the process

${\text{W}}_{\text{74}}^{\text{184}}\to {\text{74}}_{\text{1}}^{\text{1}}{\text{p +110}}_{\text{0}}^{\text{1}}\text{n}$

There are 74 protons and 110 neutrons in the tungsten nucleus.

The mass of 74 proton is

$\text{74 }\times 1.00728amu=74.5387amu$

The mass of 110 neutron is

$\text{74×5}{\text{.4858×10}}^{\text{-4}}\text{amu=0}\text{.04059amu}$

$\text{110×1}\text{.008665amu=110}\text{.9532amu}$

So the predicted mass of ${\text{W}}_{\text{74}}^{\text{184}}$ is

$\text{74}\text{.5387amu+0}\text{.04059amu+110}\text{.9532amu=185}\text{.5325 amu}$

The mass defect is found

$\text{ΔE=183}\text{.950928amu-185}\text{.5325amu=-1}\text{.5816amu}$

$\text{-1}\text{.5816amu×}\frac{\text{1kg}}{\text{6}{\text{.0221418×10}}^{\text{26}}\text{amu}}{\text{=-26262×10}}^{\text{-27}}\text{kg}$

The energy change for the process is 

$\text{ΔE=(-2}{\text{.6262×10}}^{\text{-27}}\text{kg)(2}{\text{.99792458×10}}^{\text{8}}{\text{m/s)}}^{\text{2}}$

$\text{ΔE=-2}{\text{.3603×10}}^{\text{-10}}\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{=-2}{\text{.3603×10}}^{\text{-10}}\text{J}$

Thus the nuclear binding energy is $\text{2}{\text{.3603×10}}^{\text{-10}}\text{J}$.

The nuclear binding energy per nucleon is obtained as follow:

$\frac{\text{2}{\text{.3603×10}}^{\text{-10}}\text{J}}{\text{184nucleons}}\text{=1}{\text{.2828×10}}^{\text{-12}}\text{J/nucleon}$

Conclusion

The nuclear binding energy and the binding energy per nucleon for the given isotopes were calculated