Chapter 11, Problem 6C

To determine

To verify:

The Q-values for fusion of two deuterium atoms and helium - 3 nucleus.

Explanation of Solution

Given info:

Nuclear Reaction

= $\text{H}\text{+}\text{H}\text{e}\to \text{H}\text{e +}\text{p}\text{+18}\text{.3 MeV}$

Mass of Helium- 4 ( $\text{H}\text{e}$)

= $\text{4}\text{.00260 u}$

Mass of Helium- 3

= $\text{3}\text{.01603 u}$

Mass of nucleus

= $\text{2}\text{.01410 u}$

Mass of deuterium

= $\text{2}\text{.01410 u}$

Mass of neutron

= $\text{1}\text{.00896 u}$

Mass of proton

= $\text{1}\text{.00875 u}$

Speed of light in vacuum

= $3\times {10}^8\text{ m/s}$.

Formula used:

$\text{H}\text{+}\text{H}\to \text{H}\text{e +}\text{n}{\text{+Q}}_1$

The mass defect relation and its Q value from Einstein’s Mass- energy relation is given by,

$\begin{array}{l}\Delta {\text{m}}_1{\text{ = (2m}}_{\text{H}}\text{) }-{\text{ (m}}_{\text{He}}+{\text{ m}}_{\text{n}})----(1)\\ {\text{Q}}_1\text{ = }\Delta {\text{m}}_1{\text{c}}^2----(2)\end{array}$

For the second equation,

$\text{H}\text{+}\text{H}\text{e}\to \text{H}\text{e +}\text{P}{\text{+Q}}_2$

The mass defect relation and its Q value from Einstein’s Mass- energy relation is given by,

$\begin{array}{l}\Delta {\text{m}}_2{\text{ = (m}}_{\text{H}}{\text{+m}}_{\text{H}\text{e}}\text{) }-{\text{ (m}}_{\text{H}\text{e}}+{\text{ m}}_{\text{P}})----(3)\\ {\text{Q}}_2\text{ = }\Delta {\text{m}}_2{\text{c}}^2----(4)\end{array}$.

Calculation:

Substituting the given values in equation 1 and 2, we get

$\begin{array}{l}\Delta {\text{ m}}_1\text{ = (2}\times \text{2}\text{.01410 u) }-\text{ (3}\text{.01603 u + 1}\text{.00869 u)}\\ \Delta {\text{ m}}_1\text{ = 0}\text{.00348 u}\\ \Delta {\text{ m}}_1=\text{0}\text{.00348 u }\times \left(\frac{1.66\times {10}^{-27}\text{kg}}{1\text{ u}}\right)\\ \Delta {\text{ m}}_1\text{ = 5}\text{.7768}\times {\text{10}}^{-30}\text{kg}\\ {\text{Q}}_1\text{ = 5}\text{.7768}\times {\text{10}}^{-30}\text{kg }\times \text{ (3}\times {\text{10}}^8{\text{m/s)}}^2\\ {\text{Q}}_1\text{ = 5}\text{.2}\times {\text{10}}^{-13}\text{J}\\ {\text{Q}}_1\text{ = 5}\text{.2}\times {\text{10}}^{-13}\text{J }\times \left(\frac{1\text{ eV}}{1.6\times {10}^{-19}\text{ J}}\right)=3.3\times {10}^6\text{eV}\\ {\text{Q}}_1\text{ = }3.3\times {10}^6\text{eV }\times \left(\frac{1\text{ MeV}}{1\times {10}^3\text{ eV}}\right)\end{array}$

${\text{Q}}_1\text{ = }3.3\text{ MeV}$

For second equation substituting the values in 3 and 4,

$\begin{array}{l}\Delta {\text{ m}}_2\text{ = (2}\text{.01410 u + 3}\text{.01603 u) }-\text{ (4}\text{.00260 u + 1}\text{.00785 u)}\\ \Delta {\text{ m}}_2\text{ = 0}\text{.01968 u}\\ \Delta {\text{ m}}_2=\text{0}\text{.01968 u }\times \left(\frac{1.66\times {10}^{-27}\text{kg}}{1\text{ u}}\right)\\ \Delta {\text{ m}}_2\text{ = 3}\text{.26688}\times {\text{10}}^{-29}\text{kg}\\ {\text{Q}}_2\text{ = 3}\text{.26688}\times {\text{10}}^{-29}\text{kg }\times \text{ (3}\times {\text{10}}^8{\text{m/s)}}^2\\ {\text{Q}}_2\text{ = 2}\text{.941}\times {\text{10}}^{-12}\text{J}\\ {\text{Q}}_2\text{ = 2}\text{.941}\times {\text{10}}^{-12}\text{J }\times \left(\frac{1\text{ eV}}{1.6\times {10}^{-19}\text{ J}}\right)=18.3\times {10}^6\text{eV}\\ {\text{Q}}_2\text{ = 18}.3\times {10}^6\text{eV }\times \left(\frac{1\text{ MeV}}{1\times {10}^3\text{ eV}}\right)\end{array}$

${\text{Q}}_2\text{ = 18}.3\text{ MeV}$.

Conclusion:

Thus, the values for fusion of two deuterium atoms and helium - 3 nucleus is verified.