Chapter 7, Problem 1DE

Interpretation Introduction

Interpretation: The nuclear masses of the isotopes of beryllium are given and the nuclei which has largest binding energy per nucleon is to be determined.

Concept Introduction: The binding energy per nucleon is directly proportional to the loss in mass in the nucleus. As the lass in mass increases, the binding energy per nucleon increases.

To determine: The isotope which has highest binding energy per nucleon out of $\text{B}\text{e},\text{B}\text{e}$ and $\text{B}\text{e}$ .

Answer to Problem 1DE

Solution: The isotope $\text{B}\text{e}$ has largest binding energy per nucleon out of the given isotopes.

Explanation of Solution

Given The nuclear mass of $\text{B}\text{e}$ is $7.0147\text{amu}$ .
The nuclear mass of $\text{B}\text{e}$ is $9.0100\text{amu}$ .
The nuclear mass of $\text{B}\text{e}$ is $10.0113\text{amu}$ .

The loss in mass of the nucleus is the difference of sum of the masses of protons and neutrons and the nuclear mass of the nuclei.

The loss in mass of the given isotopes is calculated as,
For $\text{B}\text{e}$

$\begin{array}{c}\Delta m=\left(3\times 1.00866\text{amu}+4\times 1.00728\text{amu}\right)-7.0147\text{amu}\\ =0.04040\text{amu}\end{array}$

For $\text{B}\text{e}$

$\begin{array}{c}\Delta m=\left(5\times 1.00866\text{amu}+4\times 1.00728\text{amu}\right)-9.0100\text{amu}\\ =0.06242\text{amu}\end{array}$

For $\text{B}\text{e}$

$\begin{array}{c}\Delta m=\left(6\times 1.00866\text{amu}+4\times 1.00728\text{amu}\right)-10.0113\text{amu}\\ =0.06978\text{amu}\end{array}$

Since, the binding energy per nucleon is directly proportional to the loss in mass. Therefore, the isotope which has highest loss in mass will have highest binding energy per nucleon. Thus, the isotope $\text{B}\text{e}$ has highest binding energy per nucleon.

Conclusion

The isotope $\text{B}\text{e}$ has largest binding energy per nucleon out of the given isotopes.