Chapter 24, Problem 72A

(a)

Interpretation Introduction

Interpretation:

The mass defect of ${}_{\text{2}}^{\text{3}}\text{He}$ is to be calculated.

Concept introduction:

The mass defect is defined as the difference in mass between a nucleus and its component nucleons.

Answer to Problem 72A

$\text{Mass}\text{defect}\text{of}{}_{\text{2}}^{\text{3}}\text{He}\text{is}\text{0}\text{.005898amu}$

Explanation of Solution

, ${}_{\text{1}}^{\text{1}}\text{H}{+}_{\text{1}}^{\text{2}}\text{H}{\to }_{\text{2}}^{\text{3}}\text{He}+\text{γ}$

$\begin{array}{c}{}_{\text{1}}^{\text{1}}\text{H}=\text{1}\text{.007825amu}\\ {}_{\text{1}}^{\text{2}}\text{H}=\text{2}\text{.014102amu}\\ {}_{\text{2}}^{\text{3}}\text{He}=\text{3}\text{.016029amu}\end{array}$

Mass defect, $\text{Δm}={\text{m}}_{\text{products}}-{\text{m}}_{\text{reaction}}$

Assume, $\text{γ}$ have no mass.

Therefore, mass defect of ${}_{\text{2}}^{\text{3}}\text{He}$

$\begin{array}{c}\text{Δm}={\text{m}}_{\text{products}}-{\text{m}}_{\text{reaction}}\\ \text{Δm}=(1.007825+2.014102)-3.016029\\ \text{Δm}=3.016029-3.021927\\ \text{Δm}=-0.005898\text{amu}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The energy released in the process needs to be determined.

Concept introduction:

The energy released is calculated by formula:

$\Delta \text{E}={\text{Δmc}}^{\text{2}}$

Where,

$\begin{array}{l}\text{Δm = change in mass}\\ \text{c= speed of light}\end{array}$

Answer to Problem 72A

The energy released by the process is $1.8\times 10^{16}\text{MeV}$.

Explanation of Solution

Write the expression for 1 amu.

$1amu=1.66\times 10^{-27}kg$

Therefore,

$\begin{array}{c}\Delta m=0.005898\times 1.66\times 10^{-27}\\ =0.00979\times 10^{-27}kg\end{array}$

Energy releases,

$\begin{array}{l}\Delta \text{E}={\text{Δmc}}^{\text{2}}\\ \Delta E=(0.00979\times {10}^{-27}\text{kg}){(}^3\\ \Delta E=0.029\times {10}^{-27}\text{J}\end{array}$

Now, $1J=6.242\times 10^{12}MeV$

$\begin{array}{l}\Delta E=0.029\times {10}^{-27}\times 6.242\times 10^{12}\\ =1.8\times 10^{16}MeV\end{array}$

Thus, the energy released by the process is $1.8\times 10^{16}\text{MeV}$.