Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a stable, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings:

$$\begin{gathered} \left(a\right)\mathbf{}{}^{\mathbf{238}}\mathbf{U}\mathbf{\to }{}^{\mathbf{232}}\mathbf{Th}\mathbf{+}{}^{\mathbf{3}}\mathbf{He}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(b\right)\mathbf{}{}^{\mathbf{235}}\mathbf{U}\mathbf{\to }{}^{\mathbf{231}}\mathbf{Th}\mathbf{+}{}^{\mathbf{4}}\mathbf{He}\mathbf{}\mathbf{} \\ \left(c\right)\mathbf{}{}^{\mathbf{235}}\mathbf{U}\mathbf{\to }{}^{\mathbf{230}}\mathbf{Th}\mathbf{+}{}^{\mathbf{5}}\mathbf{He}\mathbf{}\mathbf{} \end{gathered}$$

The needed atomic masses are

role="math" localid="1661928659878" $$\begin{gathered} {}^{\mathbf{232}}\mathbf{Th}\mathbf{}\mathbf{}\mathbf{}\mathbf{232}\mathbf{.}\mathbf{0381}\mathbf{}\mathbf{}\mathbf{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{3}}\mathbf{He}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{0160}\mathbf{}\mathbf{}\mathbf{u} \\ {}^{\mathbf{231}}\mathbf{Th}\mathbf{}\mathbf{}\mathbf{}\mathbf{231}\mathbf{.}\mathbf{0363}\mathbf{}\mathbf{}\mathbf{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{4}}\mathbf{He}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0026}\mathbf{}\mathbf{}\mathbf{u} \\ {}^{\mathbf{230}}\mathbf{Th}\mathbf{}\mathbf{}\mathbf{}\mathbf{230}\mathbf{.}\mathbf{0331}\mathbf{}\mathbf{}\mathbf{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{5}}\mathbf{He}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0122}\mathbf{}\mathbf{}\mathbf{u} \\ {}^{\mathbf{235}}\mathit{U}\mathbf{}\mathbf{}\mathbf{}\mathbf{235}\mathbf{.}\mathbf{0429}\mathbf{}\mathbf{}\mathbf{u} \end{gathered}$$

The given atomic masses of the nuclides and alpha particles are:

$$\begin{gathered} {}^{232}\mathrm{Th}232.0381\mathrm{u}{}^3\mathrm{He}3.0160\mathrm{u} \\ {}^{231}\mathrm{Th}231.0363\mathrm{u}{}^4\mathrm{He}4.0026\mathrm{u} \\ {}^{230}\mathrm{Th}230.0331\mathrm{u}{}^5\mathrm{He}5.0122\mathrm{u} \\ {}^{235}\mathrm{U}235.0429\mathrm{u} \end{gathered}$$

Massive nuclides tend to undergo alpha decay releasing disintegration energy. The disintegration energy, also known as the Q-value, is the energy that is absorbed or released when a nuclear reaction takes place. The Q-value is positive if the reaction is exothermic and negative if the reaction is endothermic. The potential barrier height of the nucleus indicates the energy it needs to overcome the internal forces and become an individual nucleus from the parent nucleus.

Formula:

The disintegration energy of a nuclear reaction,

$Q=\left(m_{parentnucleus}-\left(\sum _{}{m}_{daughternuclei}\right)\right)c^2$ …… (i)

The disintegration energyfor uranium-235 “decaying” into thorium-232is given using the atomic masses and equation (i) as follows:

$\mathrm{Q}=\left({\mathrm{m}}_{{235}_{\mathrm{U}}}-{\mathrm{m}}_{{232}_{\mathrm{Th}}}-{\mathrm{m}}_{3_{\mathrm{He}}}\right){\mathrm{c}}^2$

Substitute the values and solve as:

$$\begin{gathered} Q=\left(235.0429u-232.0381u-3.0160u\right)\left(931.5\frac{MeV}{u}\right) \\ =-9.50MeV \end{gathered}$$

Hence, the disintegration energy is - 9.50 MeV.

The disintegration energyfor uranium-235 decaying into thorium-231is given using the atomic masses and equation (i) as follows:

$\mathrm{Q}=\left({\mathrm{m}}_{{235}_{\mathrm{U}}}-{\mathrm{m}}_{{231}_{\mathrm{Th}}}-{\mathrm{m}}_{4_{\mathrm{He}}}\right){\mathrm{c}}^2$

Substitute the values and solve as:

$$\begin{gathered} Q=\left(235.0429u-231.0363u-4.0026u\right)\left(931.5\frac{MeV}{u}\right) \\ =4.66MeV \end{gathered}$$

Hence, the disintegration energy is 4.66 MeV.

The disintegration energyfor uranium-235 decaying into thorium-230is given using the atomic masses and equation (i) as follows:

$\mathrm{Q}=\left({\mathrm{m}}_{{235}_{\mathrm{U}}}-{\mathrm{m}}_{{230}_{\mathrm{Th}}}-{\mathrm{m}}_{5_{\mathrm{He}}}\right){\mathrm{c}}^2$

Substitute the values and solve as:

role="math" localid="1661929289112" $$\begin{gathered} Q=\left(235.0429u-230.0331u-5.0122u\right)\left(931.5\frac{MeV}{u}\right) \\ =-1.30MeV \end{gathered}$$

Hence, the disintegration energy is - 1.30 MeV.

Only the second decay process (the $\alpha$decay) is spontaneous, as it releases energy considering the positive sign of Q-value.