Chapter 21, Problem 21.109QA

Interpretation Introduction

To verify a statement:

From Section 21.6 of the textbook read this statement -  “no energy would be released if two $He$ nuclei were to fuse together to form $Be$. Similarly, $Be$ require no energy to spontaneously decompose into two $He$ nuclei, so they would immediately do so.”

Verify this statement by calculating the binding energy of $Be$  and compare it to that of $He$

Answer to Problem 21.109QA

Solution:

Binding energy for $Be\mathrm{ }=9.844\times {10}^{-12} J$

Binding energy for 2 atoms of $He\mathrm{ }=9.067 \times {10}^{-12}J$

The binding energies of $He$ and $Be$ are very close to each other. Therefore, $Be$ nuclei require no energy to spontaneously decompose into $He$ nuclei.

Explanation of Solution

1) Concept:

To calculate the binding energy of the nuclei of two isotopes, we need to calculate the mass defect of the nuclei. From the exact mass of the isotopes, we will calculate the exact mass of each isotope. From the mass of each isotope, we will calculate the mass of the nucleons. Calculate the mass of the nucleus by using the mass of photons and neutrons. Calculate the mass defect from the mass of nucleus and mass of nucleons. We will calculate the binding energy from mass defect.

2) Formula:

i)  $∆m=mass of nucleus-mass of nucleons$     

ii) $E= {∆mc}^2$

3) Given:

i) Mass of  $He=4.000 amu$     

ii) Mass of $Be=8.000 amu$

iii) Mass of single neutron is $1.67493 \times {10}^{-27}kg$

iv) Mass of single proton is $1.67262 \times {10}^{-27}kg$

v) Mass of electrons is $9.10939\times {10}^{-31}kg$

4) Calculations:

First we convert the exact mass of one atom of each isotope from $amu$ to kg:

$He :4.0026 amu \times \frac{1.66054 \times {10}^{-27}kg}{amu }=6.64648\times {10}^{-27}kg$

$Be: 8.00 amu \times \frac{1.66054 \times {10}^{-27}kg}{amu }=1.328432\times {10}^{-26}kg$

Since we are comparing the masses of $He$ and $Be$ nuclei, we need to subtract the mass of 2 and 4 electrons from the exact masses of the $He$ and $Be$ atoms.

$He : 6.64648\times {10}^{-27}kg-2\left(9.10939\times {10}^{-31}kg\right)=6.64466\times {10}^{-27}kg$

$Be: 1.328432\times {10}^{-26}kg-4\left(9.10939\times {10}^{-31}kg\right)=1.32807\times {10}^{-26}kg$

Next we calculate the mass of the protons and neutrons in the nucleus of one atom of each isotope:

$He :\left( 2 protons \times \frac{1.67262 \times {10}^{-27}kg}{proton}\right)+\left( 2 neutrons\times \frac{1.67493 \times {10}^{-27}kg}{neutron}\right)$

$=6.69510 \times {10}^{-27}kg$

$Be: \left( 4 protons \times \frac{1.67262 \times {10}^{-27}kg}{proton}\right)+\left( 4 neutrons\times \frac{1.67493 \times {10}^{-27}kg}{neutron}\right)$

$=1.33902 \times {10}^{-26}kg$

The difference between the exact mass of the nucleus of the isotope and its subatomic particles is

 $∆m=mass of nucleus-mass of nucleons$    

 

$He: ∆m= 6.69510 \times {10}^{-27}kg-6.64466\times {10}^{-27}kg$

$=5.044 \times {10}^{-29}kg$

$Be: ∆m= 1.33902 \times {10}^{-26}kg- 1.32807\times {10}^{-26}kg$

$=1.09524\times {10}^{-28}kg$

Calculation of the binding energy corresponding to this difference in mass:

$E= {mc}^2$

$He:E=\left(5.044 \times {10}^{-29}kg \right)\times {\left(2.998 \times {10}^8\frac{m}{s^2}\right)}^2=4.5335\times {10}^{-12}kg. \frac{m^2}{s^2}$

$=4.5335\times {10}^{-12} J$

It is the binding energy of just a single helium nucleus. The binding energy for two helium nuclei is

$=4.5335\times {10}^{-12} J+4.5335\times {10}^{-12} J=9.067\times {10}^{-12} J$

$Be:E= \left(1.09524\times {10}^{-28}kg\right) {\left(2.998 \times {10}^8\frac{m}{s^2}\right)}^2= 9.844\times {10}^{-12} kg. \frac{m^2}{s^2}$

$=9.844\times {10}^{-12} J$

The binding energies of $He$ and $Be$ are very close to each other. Therefore, $Be$ nuclei require no energy to spontaneously decompose into $He$ nuclei.

Conclusion:

We calculated the binding energy of each isotope from the exact mass and mass defect.