Chapter 20, Problem 20.81QP

Interpretation Introduction

Interpretation:

• Change in energy for given nuclear reaction has to be calculated.
• Energy change (in MeV) per ${}_{\text{1}}^{\text{2}}\text{H}$ nucleus has to be calculated.

Answer to Problem 20.81QP

• Change in energy for given nuclear reaction is $\text{-1}{\text{.688×10}}^{\text{12}}\text{J}$
• Energy change (in MeV) per ${}_{\text{1}}^{\text{2}}\text{H}$ nucleus is $\text{-17}\text{.501 MeV}$

Explanation of Solution

To determine: Nuclear masses of each nucleus

$\begin{array}{l}{\text{Nuclear mass of }}_1^{\text{2}}\text{H = 2}\text{.01400 amu - 0}\text{.000549amu}\\ \text{= 2}\text{.013451}\text{amu}\end{array}$

$\begin{array}{l}{\text{Nuclear mass of }}_{\text{1}}^{\text{3}}\text{H = 3}\text{.01605 amu - 0}\text{.000549amu}\\ \text{= 3}\text{.015501}\text{amu}\end{array}$

Write the suitable nuclear mass below each nuclide symbol.  After calculate change in mass $\text{(Δm)}$

$\begin{array}{l}{}_{\text{1}}^{\text{2}}\text{H}\text{+}{}_{\text{1}}^{\text{3}}\text{H}\stackrel{}{\to }{}_{\text{2}}^{\text{4}}\text{He}{\text{+}}_{\text{0}}^{\text{1}}\text{n}\\ \text{2}\text{.013451}\text{3}\text{.015501}\text{4}\text{.001502}\text{1}\text{.008665amu}\end{array}$

The change in mass for molar quantity is,

$\begin{array}{l}{\text{ΔE= (Δm)c}}^{\text{2}}\text{= (-0}{\text{.018785×10}}^{\text{-3}}\text{kg)(2}{\text{.998×10}}^{\text{8}}{\text{m/s)}}^{\text{2}}\\ \\ \text{= -1}\text{.68839}\times {\text{10}}^{\text{12}}\text{kg}•{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \\ \text{= -1}{\text{.688×10}}^{\text{-12}}\text{J/mol}\end{array}$

To determine: energy change in MeV for one ${}_{\text{1}}^{\text{2}}\text{H}$ nucleus

Finally, calculate the energy change in MeV for one ${}_{\text{1}}^{\text{2}}\text{H}$ nucleus

$\begin{array}{l}\frac{\text{-1}\text{.68839×1012J}}{\text{1 mol }}\text{× }\frac{\text{1 mol}}{\text{6}{\text{.022×10}}^{\text{23}}\text{ nuclei }}\text{× }\frac{\text{1 MeV}}{\text{1}{\text{.602×10}}^{\text{-13}}\text{ J }}\text{= -17}\text{.501 }\\ \\ \text{= 17}\text{.50 MeV}\end{array}$

Conclusion

• Change in energy for given nuclear reaction was calculated as $\text{-1}{\text{.688×10}}^{\text{12}}\text{J}$
• Energy change (in MeV) per ${}_{\text{1}}^{\text{2}}\text{H}$ nucleus was calculated as $\text{-17}\text{.501 MeV}$