31.4 Nuclear decay and conservation laws  (Page 3/27)

Alpha decay energy found from nuclear masses

Find the energy emitted in the $\alpha$ decay of ${}^{\text{239}}\text{Pu}$ .

Strategy

Nuclear reaction energy, such as released in α decay, can be found using the equation $E=(\mathrm{\Delta }m)c^2$ . We must first find $\mathrm{\Delta }m$ , the difference in mass between the parent nucleus and the products of the decay. This is easily done using masses given in [link] .

Solution

The decay equation was given earlier for ${}^{\text{239}}\text{Pu}$ ; it is

Thus the pertinent masses are those of ${}^{\text{239}}\text{Pu}$ , ${}^{\text{235}}\text{U}$ , and the $\alpha$ particle or ${}^4\text{He}$ , all of which are listed in [link] . The initial mass was $m{(}^{\text{239}}\text{Pu})=\text{239}\text{.}\text{052157 u}$ . The final mass is the sum $m{(}^{\text{235}}\text{U})\text{+}m{(}^4\text{He})\text{= 235}\text{.}\text{043924 u + 4.002602 u = 239.046526 u}$ . Thus,

Now we can find $E$ by entering $\mathrm{\Delta }m$ into the equation:

We know $\text{1 u}=\text{931.5 MeV/}{c}^2$ , and so

Discussion

The energy released in this $\alpha$ decay is in the $\text{MeV}$ range, about ${\text{10}}^6$ times as great as typical chemical reaction energies, consistent with many previous discussions. Most of this energy becomes kinetic energy of the $\alpha$ particle (or ${}^4\text{He}$ nucleus), which moves away at high speed. The energy carried away by the recoil of the ${}^{\text{235}}\text{U}$ nucleus is much smaller in order to conserve momentum. The ${}^{\text{235}}\text{U}$ nucleus can be left in an excited state to later emit photons ( $\gamma$ rays). This decay is spontaneous and releases energy, because the products have less mass than the parent nucleus. The question of why the products have less mass will be discussed in [link] . Note that the masses given in [link] are atomic masses of neutral atoms, including their electrons. The mass of the electrons is the same before and after $\alpha$ decay, and so their masses subtract out when finding $\mathrm{\Delta }m$ . In this case, there are 94 electrons before and after the decay.