Chapter 20, Problem 20.122QP

(a)

Interpretation Introduction

Interpretation:

Sodium isotope is only occurring in nature which is does not radioactive.

Mass defect of Sodium has to be calculated.

Answer to Problem 20.122QP

Mass defect of Sodium is $\text{0}\text{.2003297 amu}$.

Explanation of Solution

To calculate: Mass defect of Sodium

Atomic mass of $\text{Na}$= $\text{22}\text{.99 amu}$

$\begin{array}{l}\text{Mass of eleven protons =11×1}\text{.00728 amu}\\ \text{= 11}\text{.08008 amu}\\ \\ \text{Mass oftwelveneutrons=12×1}\text{.008665 amu}\\ \text{=12}\text{.10398 amu}\\ \\ \text{Total mass of nucleons =(11}\text{.08008+12}\text{.10398) amu}\\ \text{=22}\text{.98373029 amu}\end{array}$

$\begin{array}{l}{\text{Nuclear mass of }}_{\text{11}}^{\text{23}}\text{Na = 22}\text{.98976929 amu - (11× 0}\text{.000549 amu)}\\ \text{= 22}\text{.98373029 amu}\end{array}$

$\begin{array}{l}\text{Mass defect = total mass - nuclear mass}\\ \text{=(23}\text{.18406-22}\text{.98373029)amu}\\ \text{= 0}\text{.2003297amu}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Sodium isotope is only occurring in nature which is does not radioactive.

The nuclear binding energy of Sodium nucleon in $\text{MeV}$ has to be calculated.

Answer to Problem 20.122QP

The nuclear binding energy of Sodium nucleon is $\text{186}\text{.64 MeV}$.

Explanation of Solution

To calculate: nuclear binding energy of Sodium nucleon

$\begin{array}{l}{\text{ΔE = (Δm)c}}^{\text{2}}\text{ = 0}\text{.2003297amu ×}\frac{\text{1g}}{\text{6}{\text{.022×10}}^{\text{23}}\text{×1amu}}\text{×}\frac{\text{1kg}}{{\text{10}}^{\text{3}}\text{g}}\\ \\ \text{×(2}{\text{.998×108m/s)}}^{\text{2}}\text{×}\frac{\text{1MeV}}{\text{1}{\text{.602×10}}^{\text{-13}}\text{J}}\\ \\ \text{=186}\text{.64 MeV}\\ \\ \text{Binding energy per nucleon =}\frac{\text{186}\text{.64 MeV}}{\text{23nucleons}}\\ \\ \text{=8}\text{.115MeV/nucleon}\end{array}$