Chapter 20, Problem 20.82QP

Interpretation Introduction

Interpretation:

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

Answer to Problem 20.82QP

• Change in energy for given nuclear reaction is $\text{-5}{\text{.0×10}}^{\text{10}}\text{J}$
• Energy change (in MeV) per ${}_{\text{1}}^{\text{2}}\text{H}$ nucleus is $\text{-0}\text{.52 MeV}$

Explanation of Solution

To determine: Nuclear mass of ${}_{\text{1}}^1\text{H}$ nucleus

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

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

$\begin{array}{l}{}_{\text{1}}^{\text{1}}\text{H}\text{+}{}_{\text{1}}^{\text{1}}\text{H}\stackrel{}{\to }{}_{\text{1}}^{\text{2}}\text{H}{\text{+}}_{\text{1}}^{\text{0}}\text{e}\\ \text{1}\text{.00728}\text{1}\text{.00728}\text{2}\text{.013451}\text{0}\text{.000549 amu}\end{array}$

$\begin{array}{l}\text{Δm =}\left[\text{2}\text{.013451 + 0}\text{.000549 - (2×1}\text{.00728)}\right]\text{amu}\\ \\ \text{= -0}\text{.000560}\text{amu}\end{array}$

The change in mass for molar quantity is,

$\begin{array}{l}{\text{ΔE= (Δm)c}}^{\text{2}}\text{= (-5}{\text{.6×10}}^{\text{-7}}\text{kg)(3}{\text{.00×10}}^{\text{8}}{\text{m/s)}}^{\text{2}}\\ \\ \text{= -5}\text{.04}\times {\text{10}}^{\text{10}}\text{kg}•{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \\ \text{= -5}{\text{.0×10}}^{\text{-10}}\text{J}\end{array}$

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

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

$\begin{array}{l}\frac{\text{-5}{\text{.04×10}}^{\text{10}}\text{J}}{\text{1 mol }}\text{× }\frac{\text{1 mol}}{\text{6}{\text{.022×10}}^{\text{23}}\text{ nuclei }}\text{× }\frac{\text{1 MeV}}{\text{1}{\text{.602×10}}^{\text{-13}}\text{ J }}\text{= -0}\text{.522 }\\ \\ \text{= -0}\text{.52 MeV}\end{array}$

Conclusion

• Change in energy for given nuclear reaction was calculated as $\text{-5}{\text{.0×10}}^{\text{10}}\text{J}$
• Energy change (in MeV) per ${}_{\text{1}}^{\text{2}}\text{H}$ nucleus was calculated as $\text{-0}\text{.52 MeV}$